ICSE Class 7 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 7 · 69 solved questions
In the adjoining figure:
(i) Name the vertex opposite to side PQ.
(ii) Name the side opposite to vertex Q.
(iii) Name the angle opposite to side QR.
(iv) Name the side opposite to ∠R.
(i) The vertex opposite to side PQ is R.
(ii) The side opposite to vertex Q is PR.
(iii) The angle opposite to side QR is ∠P.
(iv) The side opposite to ∠R is PQ.
Look at the figures given below and classify each of the triangle according to its (a) sides (b) angles (You may judge the nature of the angle by observation):
(a) Classification according to sides:
(i) Isosceles triangle (two sides equal)
(ii) Scalene triangle (all sides unequal)
(iii) Equilateral triangle (all three sides equal)
(iv) Isosceles triangle (two sides equal)
(v) Scalene triangle (all sides unequal)
(vi) Isosceles triangle (two sides equal)
(b) Classification according to angles:
(i) Acute angled triangle (all angles less than 90°)
(ii) Right angled triangle (one angle equal to 90°)
(iii) Acute angled triangle (all angles less than 90°)
(iv) Obtuse angled triangle (one angle greater than 90°)
(v) Obtuse angled triangle (one angle greater than 90°)
(vi) Right angled triangle (one angle equal to 90°)
In the adjoining ΔPQR, if D is the mid-point of $\overline{QR}$, then
(i) $\overline{PM}$ is .....
(ii) $\overline{PD}$ is .....
Is QM = MR?
(i) Since PM is the perpendicular drawn from vertex P to the opposite side QR, $\overline{PM}$ is an altitude.
(ii) Since D is the mid-point of QR and PD is drawn from vertex P to this mid-point, $\overline{PD}$ is a median.
M is the foot of the altitude (the point where the perpendicular meets QR), while D is the mid-point of QR. These two points are not the same, so M does not divide QR into two equal parts.
Hence, QM ≠ MR.
Will an altitude always lie in the interior of a triangle? If no, draw a rough sketch to show such a case.
No, an altitude does not always lie in the interior of a triangle.
In an obtuse angled triangle, the altitude drawn from a vertex of an acute angle falls outside the triangle, i.e. on the extension of the opposite side. Hence such an altitude lies in the exterior of the triangle.
Can you think of a triangle in which two altitudes of the triangle are its sides? If yes, draw a rough sketch to show such a case.
Yes — a right angled triangle.
In a right angled triangle, the two sides containing the right angle (the legs) are perpendicular to each other. Therefore each leg is the altitude corresponding to the other leg. Thus two of the altitudes of the triangle are its own sides.
Draw rough sketches for the following:
(i) In ΔABC, BE is a median of the triangle.
(ii) In ΔPQR, PQ and PR are altitudes of the triangle.
(iii) In ΔXYZ, YL is an altitude in the exterior of the triangle.
(i) BE is a median of ΔABC, so E is the mid-point of the opposite side AC, i.e. AE = EC.
(ii) In ΔPQR, PQ and PR are altitudes. This is possible only when PQ ⊥ PR, i.e. the triangle is right angled at P.
Here, PQ is the altitude from Q to the side PR, and PR is the altitude from R to the side PQ.
(iii) In ΔXYZ, YL is an altitude lying in the exterior of the triangle. This happens when the triangle is obtuse angled, so that the foot of the perpendicular L lies on the extension of side XZ.
Take an equilateral triangle and draw its medians and altitudes and check that the medians and altitudes are same.
Consider an equilateral triangle ABC in which AB = BC = CA.
Fold the triangle so that vertex B falls exactly on vertex C. The crease passes through vertex A and meets BC at a point D.
(1) Since B falls exactly on C, the part DB falls exactly on DC.
∴ BD = DC, i.e. D is the mid-point of BC, so AD is the median from A.
(2) On folding, ∠ADB falls exactly on ∠ADC.
∴ ∠ADB = ∠ADC
But ∠ADB + ∠ADC = 180° (linear pair)
⇒ ∠ADB + ∠ADB = 180°
⇒ 2∠ADB = 180°
⇒ ∠ADB = 90°
∴ AD ⊥ BC, so AD is also the altitude from A.
Thus, the median and the altitude from A are the same line segment AD. This can also be checked by measuring : BD and DC are found to be equal, and ∠ADB is found to be 90°.
In the same way, by folding A onto C and then A onto B, the median and the altitude from B are the same, and the median and the altitude from C are the same.
Hence, in an equilateral triangle the medians and the altitudes are the same.
Find the value of the unknown exterior angle x in each of the following diagrams:
We use the exterior angle property: an exterior angle of a triangle is equal to the sum of its two interior opposite angles.
(i) Solving,
⇒ x = 45° + 65° = 110°.
Hence, x = 110°.
(ii) Solving,
⇒ x = 55° + 40° = 95°.
Hence, x = 95°.
(iii) Solving,
⇒ x = 50° + 50° = 100°.
Hence, x = 100°.
Find the value of the unknown interior angle x in each of the following diagrams:
We use the exterior angle property: an exterior angle of a triangle is equal to the sum of its two interior opposite angles.
(i) Solving,
⇒ 115° = 50° + x
⇒ x = 115° − 50° = 65°.
Hence, x = 65°.
(ii) Solving,
⇒ 80° = 30° + x
⇒ x = 80° − 30° = 50°.
Hence, x = 50°.
(iii) Solving,
⇒ 70° = x + 36°
⇒ x = 70° − 36° = 34°.
Hence, x = 34°.
Find the value of x in each of the following diagrams:
(i) Solving,
The exterior angle 105° equals the sum of the two interior opposite angles 2x and x.
⇒ 2x + x = 105°
⇒ 3x = 105°
⇒ x = 35°
Hence, x = 35°.
(ii) Solving,
The exterior angle 125° equals the sum of the two interior opposite angles 2x and 3x.
⇒ 2x + 3x = 125°
⇒ 5x = 125°
⇒ x = 25°
Hence, x = 25°.
(iii) Solving,
By the angle sum property of a triangle,
⇒ x + 60° + 50° = 180°
⇒ x + 110° = 180°
⇒ x = 70°
Hence, x = 70°.
Find the value of x in each of the following diagrams:
(i) Solving,
The triangle is isosceles with the two base angles equal to x and the vertical angle 50°. By the angle sum property,
⇒ 50° + x + x = 180°
⇒ 2x = 130°
⇒ x = 65°
Hence, x = 65°.
(ii) Solving,
We will find both interior angles by linear pair i.e sum of all angles on a straight line = 180°.
⇒ Two interior angles are:
180° - 110° = 70°
180° - 120° = 60°
Hence, the two interior angles of the triangle are 70° and 60°.
By the angle sum property,
⇒ x + 70° + 60° = 180°
⇒ x + 130° = 180°
⇒ x = 50°
Hence, x = 50°.
(iii) Solving,
One angle of the triangle is a right angle 90° and the other is 2x, with x being the third angle. By the angle sum property,
⇒ x + 2x + 90° = 180°
⇒ 3x = 90°
⇒ x = 30°
Hence, x = 30°.
Find the values of x and y in each of the following diagrams:
(i) Solving,
y and 120° form a linear pair, so
⇒ y = 180° − 120° = 60°.
In the triangle, by the angle sum property,
⇒ x + 50° + y = 180°
⇒ x + 50° + 60° = 180°
⇒ x = 70°
Hence, x = 70° and y = 60°.
(ii) Solving,
x and the given 60° are vertically opposite angles, so
⇒ x = 60°.
In the triangle, by the angle sum property,
⇒ 40° + x + y = 180°
⇒ 40° + 60° + y = 180°
⇒ y = 80°
Hence, x = 60° and y = 80°.
(iii) Solving,
y and the given 90° are vertically opposite angles, so
⇒ y = 90°.
In the triangle, the two base angles are each x and the third angle is y = 90°. By the angle sum property,
⇒ x + x + 90° = 180°
⇒ 2x = 90°
⇒ x = 45°
Hence, x = 45° and y = 90°.
Find the values of x and y in each of the following diagrams:
(i) Here x and y are vertically opposite angles formed by the two crossing lines, so x = y. The two base angles of the triangle are each equal to x, and the vertical angle of the triangle is y. By the angle sum property,
⇒ x + x + y = 180°
⇒ x + x + x = 180°
⇒ 3x = 180°
⇒ x = 60°
∴ y = x = 60°.
Hence, x = 60° and y = 60°.
(ii) The exterior angle at the left base vertex is 125° and both the base angles of the triangle are marked x.
Since x and 125° form a linear pair,
⇒ x + 125° = 180°
⇒ x = 180° − 125° = 55°
So, each base angle of the triangle = x = 55°.
In the triangle, by the angle sum property,
⇒ y + x + x = 180°
⇒ y + 55° + 55° = 180°
⇒ y = 180° − 110° = 70°
Hence, x = 55° and y = 70°.
(iii) The apex of the larger triangle is split into y (on the left) and 50° (on the right), the base-left angle is 36° and the base-right angle is 70°.
In the triangle ADC, by the angle sum property,
⇒ x + 50° + 70° = 180°
⇒ x = 60°
Now ∠ADB and ∠ADC form a linear pair, so ∠ADB = 180° − 60° = 120°.
In the triangle ADB, by the angle sum property,
⇒ y + 36° + 120° = 180°
⇒ y = 24°
Hence, x = 60° and y = 24°.
(a) In the fig. (i) given below, DE ∥ BC. Find the value of x.
(b) In the fig. (ii) given below, ∠P = 3y. Find the values of x and y.
(c) In the fig. (iii) given below, find the values of x, y and z.
(a) Solving,
∠AED and ∠CED form a linear pair, so ∠AED = 180° − 95° = 85°.
Since DE ∥ BC, ∠ADE = ∠B = 50° (corresponding angles).
In ΔADE, by the angle sum property,
⇒ x + 85° + 50° = 180°
⇒ x = 45°
Hence, x = 45°.
(b) Solving,
In ΔPQS, ∠P = 3y, ∠PQS = y and ∠PSQ = 100°. By the angle sum property,
⇒ 3y + y + 100° = 180°
⇒ 4y = 80°
⇒ y = 20°
Now ∠QSR and ∠PSQ form a linear pair, so ∠QSR = 180° − 100° = 80°.
In ΔQSR, by the angle sum property,
⇒ x + 80° + 60° = 180°
⇒ x = 40°
Hence, x = 40° and y = 20°.
(c) Solving,
In the large triangle, the apex angle x has the other two angles 80° and 52°. By the angle sum property,
⇒ x + 80° + 52° = 180°
⇒ x = 48°
z and 143° form a linear pair, so
⇒ z = 180° − 143° = 37°.
In the small triangle, by the angle sum property,
⇒ y + x + z = 180°
⇒ y + 48° + 37° = 180°
⇒ y = 95°
Hence, x = 48°, y = 95° and z = 37°.
One of the angles of a triangle measures 80° and the other two angles are equal. Find the measure of each of the equal angles.
Let each of the two equal angles be x.
By the angle sum property of a triangle,
⇒ 80° + x + x = 180°
⇒ 2x = 100°
⇒ x = 50°
Hence, each of the equal angles measures 50°.
If one angle of a triangle is 60° and the other two angles are in the ratio 2 : 3, find these angles.
Let the other two angles be 2x and 3x.
By the angle sum property of a triangle,
⇒ 60° + 2x + 3x = 180°
⇒ 5x = 120°
⇒ x = 24°
∴ The two angles are 2x = 2 × 24° = 48° and 3x = 3 × 24° = 72°.
Hence, the two angles are 48° and 72°.
If the angles of a triangle are in the ratio 1 : 2 : 3, find the angles. Classify the triangle in two different ways.
Let the three angles be x, 2x and 3x.
By the angle sum property of a triangle,
⇒ x + 2x + 3x = 180°
⇒ 6x = 180°
⇒ x = 30°
∴ The angles are x = 30°, 2x = 60° and 3x = 90°.
Since all three angles are different, the triangle is scalene. Since one angle is 90°, it is right angled.
Hence, the angles are 30°, 60° and 90°; the triangle is a scalene triangle and a right angled triangle.
Can a triangle have three angles whose measures are
(i) 65°, 74°, 39°?
(ii) $\frac{1}{3}$ right angle, 1 right angle, 60°?
A triangle is possible only if the sum of its three angles is 180°.
(i) Solving,
⇒ 65° + 74° + 39° = 178° ≠ 180°.
Since the sum is not 180°, such a triangle is not possible.
Hence, a triangle cannot have angles 65°, 74° and 39°.
(ii) Solving,
$\frac{1}{3} \text { right angle } = \frac{1}{3} × 90° = 30°$ and 1 right angle = 90°.
⇒ 30° + 90° + 60° = 180°.
Since the sum is 180°, such a triangle is possible.
Hence, a triangle can have angles $\bm{\frac{1}{3}}$ right angle, 1 right angle and 60°.
Find the value of x in each of the following figures:
(i) Solving,
In ΔABC, AB = AC, so the angles opposite to the equal sides are equal.
⇒ ∠ABC = ∠ACB
⇒ x = 50°
Hence, x = 50°.
(ii) Solving,
In ΔABC, AB = AC, so the angles opposite to the equal sides are equal.
⇒ ∠ABC = ∠ACB = x
Since CA is produced to D,
⇒ ∠BAC + ∠DAB = 180° (linear pair)
⇒ ∠BAC + 110° = 180°
⇒ ∠BAC = 180° − 110° = 70°
In ΔABC, by angle sum property,
⇒ ∠BAC + ∠ABC + ∠ACB = 180°
⇒ 70° + x + x = 180°
⇒ 2x = 180° − 70° = 110°
⇒ x = 55°
Hence, x = 55°.
(iii) Solving,
AB is produced to D and CB is produced to E.
⇒ ∠ABC = ∠DBE = 36° (Vertically opposite angles are equal)
In ΔABC, CA = CB, so the angles opposite to the equal sides are equal.
⇒ ∠CAB = ∠CBA
⇒ x = 36°
Hence, x = 36°.
Find the value of x in each of the following figures:
(i) Solving,
In ΔABC, CA = CB, so the angles opposite to the equal sides are equal.
⇒ ∠CAB = ∠CBA = x
In ΔABC, by angle sum property,
⇒ ∠CAB + ∠CBA + ∠ACB = 180°
⇒ x + x + 40° = 180°
⇒ 2x = 180° - 40° = 140°
⇒ x = 70°
Hence, x = 70°.
(ii) Solving,
In ΔABC, BA = BC, so the angles opposite to the equal sides are equal.
⇒ ∠BAC = ∠BCA = 45°
In ΔABC, by angle sum property,
⇒ ∠ABC + ∠BAC + ∠BCA = 180°
⇒ x + 45° + 45° = 180°
⇒ x = 180° − 90° = 90°
Hence, x = 90°.
(iii) Solving,
The triangle is isosceles, so the base angles are equal = x .
As ray l is parallel to line segment AB, therefore angle ABC = 50°(corresponding angles)
Now, by angle sum property:
⇒ x + x + 50° = 180°
⇒ 2x = 180° - 50° = 130°
⇒ x = 65°.
Hence, x = 65°.
Find the values of x and y in each of the following figures:
(i) Solving,
The triangle is an isosceles triangle (two sides marked equal), so the base angles are equal = y.
As, ∠BCA + ∠ACD = 180°(linear pair)
⇒ ∠BCA + 120° = 180°
⇒ ∠BCA = 60°
⇒ ∠BCA = y = 60°(base angles of isosceles triangle)
Now, by angle sum property,
⇒ x + ∠BCA + y = 180°
⇒ x + 60° + 60° = 180°
⇒ x = 60°
Hence, x = 60° and y = 60°.
(ii) Solving,
The triangle is isosceles with two equal sides, so the base angles are equal.
Now, ∠ACB + ∠ACD = 180° (linear pair)
⇒ ∠ACB + 115°= 180°
⇒ ∠ACB = 65°
As, ∠ACB = x = 65° (base angles of isosceles triangle are equal)
In ΔABC, by angle sum property,
⇒ x + ∠ACB + y = 180°
⇒ 65° + 65° + y = 180°
⇒ y = 50°
Hence, x = 65° and y = 50°.
(iii) Solving,
∠A splits into 48° and y (∠A = 48° + y). Here AB = AD, so the base angles of ΔABD are equal i.e ∠ABD = x
In ΔABD, by angle sum property,
⇒ ∠BAD + ∠ABD + x = 180°
⇒ 48° + x + x = 180°
⇒ 2x = 180° - 48° = 132°
⇒ x = 66°
Also AD = DC, so ΔADC is isosceles with ∠DAC = ∠DCA = y.
In ΔABC, by angle sum property,
⇒ ∠A + ∠B + ∠C = 180°
⇒ 48° + y + 66° + y = 180°
⇒ 2y + 114° = 180°
⇒ 2y = 180° - 114° = 66°
⇒ y = 33°.
Hence, x = 66° and y = 33°.
Calculate the size of each lettered angle in the following figures:
(i) Solving,
The apex angle of ΔTBD is 32° with TD = BD. As ΔTBD is isosceles,
⇒ x = ∠DBT = ∠DTB = 32°.(base angles of isosceles triangle are equal)
∴ By angle sum property,
∠TDB + ∠TBD + ∠BTD = 180°
∠TDB + 32° + 32° = 180°
∠TDB = 180° − 32° − 32° = 116°.
Now ∠BDC and ∠TDB form a linear pair on TC, so
⇒ y = ∠BDC = 180° − 116° = 64°.
In ΔBDC, BD = BC, so the base angles are equal:
⇒ ∠BDC = ∠BCD = 64°.
∴ By angle sum property,
z = ∠DBC = 180° − 64° − 64° = 52°.
Hence, x = 32°, y = 64° and z = 52°.
(ii) Solving,
In ΔBTD, BD = TD, so the base angles are equal:
⇒ ∠DBT = ∠BTD = 35°.
∴ In ΔBTD, by angle sum property,
⇒ ∠BDT = 180° − 35° − 35° = 110°.
Now ∠BDC and ∠BDT form a linear pair on TC, so
⇒ y + ∠BDT = 180°
⇒ y = ∠BDC = 180° − 110° = 70°.
In ΔBDC, BD = BC, so the base angles are equal:
⇒ ∠BCD = ∠BDC = 70°.
In ΔBDC, by angle sum property,
⇒ ∠BCD + ∠BDC + ∠DBC = 180°
⇒ ∠DBC = 40°
Now, x, ∠DBT and ∠DBC form a linear pair on TM.
⇒ x + ∠DBT + ∠DBC = 180°
⇒ x + 35° + 40° = 180°
⇒ x = 180° - 75° = 105°
Hence, x = 105° and y = 70°.
If the angles of a triangle are in the ratio 1 : 2 : 1, find all the angles of the triangle. Classify the triangle in two different ways.
Let the three angles be x, 2x and x.
By the angle sum property of a triangle,
⇒ x + 2x + x = 180°
⇒ 4x = 180°
⇒ x = 45°
∴ The angles are x = 45°, 2x = 90° and x = 45°.
Since two angles are equal, the triangle is isosceles. Since one angle is 90°, it is right angled.
Hence, the angles are 45°, 90° and 45°; the triangle is an isosceles triangle and a right angled triangle.
In an isosceles triangle, a base angle is four times its vertical angle. Find all the angles of the triangle.
Let the vertical angle be x. Then each base angle is 4x.
In an isosceles triangle the two base angles are equal, so the angles are x, 4x and 4x.
By the angle sum property of a triangle,
⇒ x + 4x + 4x = 180°
⇒ 9x = 180°
⇒ x = 20°
∴ The vertical angle is 20° and each base angle is 4 × 20° = 80°.
Hence, the angles of the triangle are 20°, 80° and 80°.
Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 2.5 cm, 4.5 cm, 8 cm
(iii) 10.2 cm, 5.8 cm, 4.5 cm
(iv) 3.4 cm, 4.7 cm, 6.2 cm
A triangle is possible only if the sum of the lengths of any two sides is greater than the third side.
(i) Solving,
⇒ 2 + 3 = 5, which is not greater than 5.
Since the sum of two sides is not greater than the third side, a triangle is not possible.
Hence, a triangle with sides 2 cm, 3 cm and 5 cm is not possible.
(ii) Solving,
⇒ 2.5 + 4.5 = 7 < 8.
Since the sum of two sides is less than the third side, a triangle is not possible.
Hence, a triangle with sides 2.5 cm, 4.5 cm and 8 cm is not possible.
(iii) Solving,
⇒ 4.5 + 5.8 = 10.3 > 10.2.
Since the sum of two sides is greater than the third side, so a triangle is possible.
Hence, a triangle with sides 10.2 cm, 5.8 cm and 4.5 cm is possible.
(iv) Solving,
⇒ 3.4 + 4.7 = 8.1 > 6.2.
Since the sum of two sides is greater than the third side, so a triangle is possible.
Hence, a triangle with sides 3.4 cm, 4.7 cm and 6.2 cm is possible.
If the lengths of two sides of a triangle are 7 cm and 10 cm, then what can be the length of the third side?
The third side must be less than the sum and greater than the difference of the other two sides.
Sum = 10 + 7 = 17 cm and difference = 10 − 7 = 3 cm.
Hence, the third side is greater than 3 cm and less than 17 cm.
We know that in a triangle, the sum of lengths of any two sides is greater than the length of the third side. Is the sum of any angles of a triangle also greater than the third angle? If no, draw a rough sketch to show such a case.
No.
The sum of any two angles of a triangle is not always greater than the third angle.
For example, consider a triangle with angles 30°, 30° and 120°. Here the sum of two angles 30° + 30° = 60°, which is less than the third angle 120°, i.e. 60° < 120°.
This is an obtuse angled triangle.
PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Since the triangle is right angled at P, QR is the hypotenuse. By the Pythagoras theorem,
⇒ QR2 = PQ2 + PR2
⇒ QR2 = 102 + 242
⇒ QR2 = 100 + 576
⇒ QR2 = 676
⇒ QR = $\sqrt{676}$ = 26
Hence, QR = 26 cm.
ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Since the triangle is right angled at C, AB is the hypotenuse. By the Pythagoras theorem,
⇒ AB2 = AC2 + BC2
⇒ 252 = 72 + BC2
⇒ 625 = 49 + BC2
⇒ BC2 = 576
⇒ BC = $\sqrt{576}$ = 24
Hence, BC = 24 cm.
Find the value of x in each of the following figures. All measurements are in centimetres.
(i) Solving,
In right angled ΔABC, ∠B = 90°, AC = 29 cm (hypotenuse), BC = 21 cm and AB = x cm.
By the Pythagoras theorem,
⇒ AC2 = AB2 + BC2
⇒ (29)2 = x2 + (21)2
⇒ x2 = (29)2 − (21)2
⇒ x2 = 841 − 441
⇒ x2 = 400
⇒ x = $\sqrt{400}$ = 20
Hence, x = 20.
(ii) Solving,
In ΔABC, AB = AC = 37 cm, AD ⊥ BC, AD = 12 cm and BC = x cm.
In right angled ΔADB, by the Pythagoras theorem,
⇒ AB2 = AD2 + BD2
⇒ (37)2 = (12)2 + BD2
⇒ BD2 = 1369 − 144 = 1225
⇒ BD = $\sqrt{1225}$ = 35 cm
In right angled ΔADC, by the Pythagoras theorem,
⇒ AC2 = AD2 + DC2
⇒ (37)2 = (12)2 + DC2
⇒ DC2 = 1369 − 144 = 1225
⇒ DC = $\sqrt{1225}$ = 35 cm
∴ x = BC = BD + DC = 35 + 35 = 70
Hence, x = 70.
(iii) Solving,
The base x = BC + CD. By the Pythagoras theorem in each right triangle △ABC and △EDC,
In △ABC,
BC $= \sqrt{13^2 − 12^2} = \sqrt{169 − 144} = \sqrt{25} = 5$
In △EDC,
CD $= \sqrt{10^2 − 6^2} = \sqrt{100 − 36} = \sqrt{64} = 8$
∴ x = 5 + 8 = 13.
Hence, x = 13.
Which of the following can be the sides of a right angled triangle?
(i) 4 cm, 5 cm, 7 cm
(ii) 1.5 cm, 2 cm, 2.5 cm
(iii) 7 cm, 5.6 cm, 4.2 cm
In case of right angled triangles, identify the right angles.
A triangle is right angled if the square of the longest side equals the sum of the squares of the other two sides.
(i) Solving,
⇒ 42 + 52 = 16 + 25 = 41, while 72 = 49.
Since 41 ≠ 49, the triangle is not a right angled triangle.
(ii) Solving,
⇒ 1.52 + 22 = 2.25 + 4 = 6.25, and 2.52 = 6.25.
Since they are equal, it is a right angled triangle; the right angle is opposite to the side of length 2.5 cm.
(iii) Solving,
⇒ 5.62 + 4.22 = 31.36 + 17.64 = 49, and 72 = 49.
Since they are equal, it is a right angled triangle; the right angle is opposite to the side of length 7 cm.
A 15 m long ladder reaches a window 12 m high from the ground on placing it against a wall. How far is the foot of the ladder from the wall?
The ladder forms the hypotenuse, the wall is the vertical side and the distance of the foot from the wall is the horizontal side. By the Pythagoras theorem,
⇒ distance2 = 152 − 122
⇒ distance2 = 225 − 144
⇒ distance2 = 81
⇒ distance = $\sqrt{81}$ = 9
Hence, the foot of the ladder is 9 m from the wall.
Find the area and the perimeter of the rectangle whose length is 15 cm and the length of one diagonal is 17 cm.
In a rectangle the diagonal, length and breadth form a right triangle, where length is the height, breadth is the base and diagonal is the hypotenuse. By the Pythagoras theorem,
⇒ breadth2 = 172 − 152
⇒ breadth2 = 289 − 225
⇒ breadth2 = 64
⇒ breadth = $\sqrt{64}$ = 8
∴ Area = length × breadth = 15 × 8 = 120 cm2.
∴ Perimeter = 2(length + breadth) = 2(15 + 8) = 2 × 23 = 46 cm.
Hence, the area is 120 cm2 and the perimeter is 46 cm.
If the diagonals of a rhombus measure 10 cm and 24 cm, find its perimeter.
The diagonals of a rhombus bisect each other at right angles. In rhombus ABCD, AC and BD are the diagonals of length 24 cm and 10 cm. So each side is the hypotenuse of a right triangle with legs equal to half of each diagonal, i.e. OB = 5 cm and AO = 12 cm.
In right-angled triangle AOB, by pythagoras theorem,
⇒ (AB)2 = (AO)2 + (OB)2
⇒ (AB)2 = 122 + 52
⇒ (AB)2 = 144 + 25
⇒ (AB)2 = 169
⇒ AB = $\sqrt{169}$ = 13
∴ Perimeter = 4 × side = 4 × 13 = 52 cm.
Hence, the perimeter of the rhombus is 52 cm.
The side of a rhombus is 5 cm. If the length of one diagonal of the rhombus is 8 cm, then find the length of the other diagonal.
The diagonals of a rhombus bisect each other at right angles. In rhombus ABCD, AB = 5 cm, AC = 8 cm. Half of the given diagonal is OA = $\frac{8}{2} = 4$ cm, and the AB = 5 cm is the hypotenuse.
The other diagonal is BD. Let half of the other diagonal OB be a.
⇒ 52 = 42 + a2
⇒ 25 = 16 + a2
⇒ a2 = 9
⇒ a = $\sqrt{9}$ = 3
∴ The other diagonal = 2 × 3 = 6 cm.
Hence, the length of the other diagonal is 6 cm.
Fill in the blanks:
(i) A triangle has atleast .... acute angles.
(ii) A triangle cannot have more than ... right angle.
(iii) A triangle cannot have more than ... obtuse angle.
(iv) In every triangle, the sum of (interior) angles of a triangle = ... right angles.
(v) In every triangle, an exterior angle + adjacent interior angle = ... degrees.
(vi) In every triangle, an exterior angle = sum of the ... interior opposite angles.
(vii) In a right angled triangle, if one of the acute angle measures 25° then the measure of the other acute angle is ......
(viii) If one of the angles of a triangle is equal to the sum of the other two angles, then the measure of this angle is ... .
(i) A triangle has atleast two acute angles.
(ii) A triangle cannot have more than one right angle.
(iii) A triangle cannot have more than one obtuse angle.
(iv) In every triangle, the sum of (interior) angles of a triangle = 2 right angles.
(v) In every triangle, an exterior angle + adjacent interior angle = 180 degrees.
(vi) In every triangle, an exterior angle = sum of the two interior opposite angles.
(vii) The other acute angle = 90° − 25° = 65°.
(viii) If one angle equals the sum of the other two, that angle = $\frac{180°}{2}$ = 90°.
State whether the following statements are true (T) or false (F):
(i) A triangle can have all three angles with measure greater than 60°
(ii) A triangle can have all three angles with measure less than 60°
(iii) If an exterior angle of a triangle is a right angle, then each of its interior opposite angle is acute.
(iv) In a right angled triangle, the sum of two acute angles is 90°
(v) If all the three sides of a triangle are equal, then it is called a scalene triangle.
(vi) Every equilateral triangle is an isosceles triangle.
(vii) Every isosceles triangle must be an equilateral triangle.
(viii) Each acute angle of an isosceles right angled triangle measures 60°
(ix) A median of a triangle always lies inside the triangle.
(x) An altitude of a triangle always lies outside the triangle.
(xi) In a triangle, sum of squares of two sides is equal to the square of the third side.
(i) False — if all three were greater than 60°, the sum would exceed 180°.
(ii) False — if all three were less than 60°, the sum would be less than 180°.
(iii) True — the two interior opposite angles add up to 90°, so each must be acute.
(iv) True — the third angle is 90°, so the two acute angles add up to 90°.
(v) False — a triangle with all three sides equal is equilateral, not scalene.
(vi) True — an equilateral triangle has all sides equal, so it certainly has two equal sides.
(vii) False — an isosceles triangle need not have all three sides equal.
(viii) False — each acute angle of an isosceles right angled triangle is 45°.
(ix) True — a median always lies inside the triangle.
(x) False — an altitude lies outside only in an obtuse angled triangle, not always.
(xi) False — this holds only for a right angled triangle, not for every triangle.
A triangle formed by the sides of lengths 4.5 cm, 6 cm and 4.5 cm is
⇒ Two of the sides are equal (4.5 cm and 4.5 cm), so the triangle is isosceles.
Hence, option 2 is the correct option.
The number of medians in a triangle is
⇒ A triangle has three vertices, and one median is drawn from each vertex.
Hence, option 3 is the correct option.
An exterior angle of a triangle is 125°. If one of the two interior opposite angle is 55°, then the other interior opposite angle is
⇒ Other interior opposite angle = 125° − 55° = 70°.
Hence, option 1 is the correct option.
In a ΔABC, if ∠A = 40° and ∠B = 55° then ∠C is
⇒ ∠C = 180° − (40° + 55°) = 180° − 95° = 85°.
Hence, option 4 is the correct option.
If the angles of a triangle are 35°, 35° and 110°, then it is
⇒ Two angles are equal (35° each), so the two opposite sides are equal, making it an isosceles triangle.
Hence, option 1 is the correct option.
A triangle whose two angles measure 30° and 120° is
⇒ The third angle = 180° − (30° + 120°) = 30°. Two angles are equal (30° each), so the triangle is isosceles.
Hence, option 2 is the correct option.
A triangle can have two
⇒ A triangle always has at least two acute angles; it cannot have two right, two obtuse or two straight angles.
Hence, option 3 is the correct option.
A triangle whose angles measure 35°, 55° and 90° is
⇒ One of the angles is 90°, so the triangle is right angled.
Hence, option 2 is the correct option.
A triangle is not possible whose angles measure
⇒ 67° + 42° + 81° = 190° ≠ 180°, so this triangle is not possible. (The other options each sum to 180°.)
Hence, option 4 is the correct option.
If in an isosceles triangle, each of the base angle is 40°, then the triangle is
⇒ Vertical angle = 180° − (40° + 40°) = 100°, which is obtuse, so the triangle is obtuse angled.
Hence, option 3 is the correct option.
A triangle is not possible with sides of lengths (in cm)
⇒ For 6, 4, 10: 6 + 4 = 10, which is not greater than 10, so a triangle is not possible.
Hence, option 1 is the correct option.
Which of the following can be the length of the third side of a triangle whose two sides measure 18 cm and 14 cm?
⇒ The third side must lie between 18 − 14 = 4 cm and 18 + 14 = 32 cm, i.e. greater than 4 cm and less than 32 cm. Among the options, only 5 cm satisfies this.
Hence, option 4 is the correct option.
In a right angled triangle, the lengths of two legs are 6 cm and 8 cm. The length of the hypotenuse is
⇒ Hypotenuse $= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
Hence, option 2 is the correct option.
If the dimensions of a rectangle are 15 m and 8 m, then the length of a diagonal is
⇒ Diagonal $= \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ m.
Hence, option 3 is the correct option.
If p, q and r are the lengths of the three sides of a triangle, then which of the following statements is correct?
⇒ By the triangle inequality, the sum of any two sides is greater than the third side, so p + q > r.
Hence, option 3 is the correct option.
Statement I: 3 cm, 4 cm and 7 cm are possible dimensions of a scalene triangle.
Statement II: The sum of the squares of two sides of a triangle is less than the square of the third side.
Statement I: 3 + 4 = 7, which is not greater than 7, so these cannot form a triangle. Statement I is false.
Statement II: This is not a general property of triangles, so Statement II is false.
Hence, option 4 is the correct option.
Statement I: A triangle with two right angles is an isosceles right angled triangle.
Statement II: An isosceles triangle has two equal angles.
Statement I: A triangle cannot have two right angles, since their sum alone would be 180°. So Statement I is false.
Statement II: An isosceles triangle does have two equal angles (the base angles). So Statement II is true.
Hence, option 2 is the correct option.
Statement I: An exterior angle of a triangle is equal to the sum of its two interior opposite angles.
Statement II: It is impossible for a triangle to have three acute angles.
Statement I: This is the exterior angle property of a triangle, which is true.
Statement II: A triangle can have three acute angles (an acute angled triangle, e.g. 60°, 60°, 60°), so this statement is false.
Hence, option 1 is the correct option.
Statement I: In a right angled triangle ΔABC with ∠C = 90°, we can say that ∠A and ∠B are acute angles.
Statement II: The hypotenuse is the longest side in a right angled triangle.
Statement I: Since ∠C = 90°, the other two angles add up to 90°, so each of ∠A and ∠B is acute. This statement is true.
Statement II: The hypotenuse lies opposite the right angle and is the longest side. This statement is also true.
Hence, option 3 is the correct option.
Find the value of x in each of the following diagrams:
(i) Solving,
The exterior angle 115° equals the sum of the two interior opposite angles (2x − 5) and x.
⇒ (2x − 5) + x = 115°
⇒ 3x − 5 = 115°
⇒ 3x = 120°
⇒ x = 40°
Hence, x = 40°.
(ii) Solving,
As, AB parallel to CD,
⇒ ∠BAO = ∠CDO = 45°(alternate angles)
In triangle OCD, by angle sum property,
⇒ ∠C + ∠CDO + ∠COD = 180°
⇒ x + 45° + 80° = 180°
⇒ x = 55°
Hence, x = 55°.
(iii) Solving,
The apex of the triangle is split into 50° and x, with base angles 72° and 45°. By the angle sum property,
⇒ (50° + x) + 72° + 45° = 180°
⇒ 167° + x = 180°
⇒ x = 13°
Hence, x = 13°.
In the adjoining figure, ∠B = 70° and ∠A = 50°. If the bisector of ∠C meets AB in D, then find ∠ADC.
In ΔABC, by the angle sum property,
⇒ ∠C = 180° − (70° + 50°) = 60°
Since CD bisects ∠C, ∠ACD = $\frac{60°}{2}$ = 30°.
In ΔADC, by the angle sum property,
⇒ ∠A + ∠ACD + ∠ADC = 180°
⇒ 50° + 30° + ∠ADC = 180°
⇒ ∠ADC = 100°
Hence, ∠ADC = 100°.
(a) In the fig. (i) given below, find the value of x.
(b) In the fig. (ii) given below, find the value of x + y.
(c) In the fig. (iii) given below, PQ = PS, find the value of x.
(a) Solving,
Given, ∠T = 20°, ∠B = 60° and ∠C = 45°
In ΔTDC, by using angle sum property,
∠T + ∠C + ∠TDC = 180°
20° + 45° + ∠TDC = 180°
∠TDC = 180° - 65° = 115°
Now, ∠TDC + ∠BDP = 180°(linear pair)
⇒ ∠BDP = 180° - 115° = 65°
In ΔPBD, by using angle sum property,
∠B + ∠BDP + ∠BPD = 180°
60° + 65° + ∠BPD = 180°
∠BPD = 180° - 125° = 55°
As, ∠TPB + ∠BPD = 180°(linear pair)
⇒ x = 180° - 55° = 125°
Hence, x = 125°.
(b) Solving,
Given, ∠C = 70°, ∠B = 95°
In ΔOCD, by angle sum property,
⇒ ∠C + ∠DOC + ∠ODC = 180°
⇒ 70° + 30° + ∠ODC = 180°
⇒ ∠ODC = 180° - 100° = 80°
As, ∠ODC + x = 180°(linear pair)
⇒ 80° + x = 180°
⇒ x = 100°
Now, ∠AOB = ∠COD = 30°
In ΔAOB, by angle sum property,
⇒ ∠B + ∠OAB + ∠BOA = 180°
⇒ 95° + ∠OAB + 30° = 180°
⇒ ∠OAB = 180° - 125° = 55°
As, ∠OAB + y = 180°(linear pair)
⇒ 55° + y = 180°
⇒ y = 125°
∴ x + y = 100° + 125° = 225°.
Hence, x + y = 225°.
(c) Solving,
In ΔPQS, PQ = PS, so the base angles are equal:
⇒ ∠PQS = ∠PSQ = 180° − 110° = 70°.
Now ∠PSR and ∠PSQ form a linear pair, so ∠PSR = 180° − 70° = 110°.
In ΔPSR, by the angle sum property,
⇒ 110° + 40° + x = 180°
⇒ x = 30°
Hence, x = 30°.
Find the values of x and y in each of the following figures:
(i) Solving,
In the triangle PMQ, PM = PQ, so the base angles are equal:
⇒ ∠PMQ = ∠PQM = 55°.
In ΔAMP, ∠A = 30° and ∠APM = y. Since ∠PMQ is the exterior angle of ΔAMP,
⇒ 55° = 30° + y
⇒ y = 25°.
The apex ∠MPQ = 180° − (55° + 55°) = 70°.
Along the straight line at P,
⇒ y + 70° + x = 180°
⇒ 25° + 70° + x = 180°
⇒ x = 85°
Hence, x = 85° and y = 25°.
(ii) Solving,
The lines l and m are parallel and AB = PB (marked equal). By alternate angles, x = ∠PAC = 66°.
Since AB = PB, ΔABP is isosceles, so ∠BAP = x = 66°.(base angles)
∴ y = ∠ABP = 180° − (66° + 66°) = 48°.
Hence, x = 66° and y = 48°.
(iii) Solving,
Here AM = MP = PB (all marked equal). In ΔAMP, AM = MP, so ∠APM = ∠A = 35°.
∴ ∠AMP = 180° − (35° + 35°) = 110°.
Now ∠PMB and ∠AMP form a linear pair, so y = ∠PMB = 180° − 110° = 70°.
In ΔPMB, MP = PB, so the base angles are equal: ∠PMB = ∠PBM = 70°.
∴ ∠MPB = 180° − (70° + 70°) = 40°.
Along the straight line at P,
⇒ 35° + 40° + x = 180°
⇒ x = 105°
Hence, x = 105° and y = 70°.
If the two acute angles of a right angled triangle are in the ratio 7 : 8, find these angles.
In a right angled triangle, the two acute angles add up to 90°. Let them be 7k and 8k.
⇒ 7k + 8k = 90°
⇒ 15k = 90°
⇒ k = 6°
∴ The angles are 7k = 7 × 6° = 42° and 8k = 8 × 6° = 48°.
Hence, the two acute angles are 42° and 48°.
If the angles of a triangle are (3x)°, (2x - 7)° and (4x - 11)°, then find the value of x.
By the angle sum property of a triangle,
⇒ 3x + (2x − 7) + (4x − 11) = 180
⇒ 9x − 18 = 180
⇒ 9x = 198
⇒ x = 22
Hence, x = 22.
In an isosceles triangle, the vertical angle is 15° greater than each of its base angles. Find all the angles of the triangle.
Let each base angle be b. Then the vertical angle is (b + 15°).
By the angle sum property of a triangle,
⇒ b + b + (b + 15°) = 180°
⇒ 3b + 15° = 180°
⇒ 3b = 165°
⇒ b = 55°
∴ Each base angle is 55° and the vertical angle is 55° + 15° = 70°.
Hence, the angles of the triangle are 55°, 55° and 70°.
Can a triangle have three sides whose lengths are
(i) 4.5 cm, 3.8 cm, 7.2 cm?
(ii) 3.2 cm, 5.3 cm, 9.4 cm?
A triangle is possible only if the sum of the lengths of any two sides is greater than the third side.
(i) Solving,
⇒ 4.5 + 3.8 = 8.3 > 7.2.
As the sum of two sides is greater than the third side, so a triangle is possible.
Hence, a triangle with sides 4.5 cm, 3.8 cm and 7.2 cm is possible.
(ii) Solving,
⇒ 3.2 + 5.3 = 8.5 < 9.4.
Since the sum of two sides is less than the third side, a triangle is not possible.
Hence, a triangle with sides 3.2 cm, 5.3 cm and 9.4 cm is not possible.
If the lengths of two sides of a triangle are 5 cm and 12 cm, then what can be the length of the third side?
The third side must be less than the sum and greater than the difference of the other two sides.
Sum = 12 + 5 = 17 cm and difference = 12 − 5 = 7 cm.
Hence, the third side is greater than 7 cm and less than 17 cm.
In the adjoining figure, all measurements are in centimetres. If AD is perpendicular to BC, find the length of AB.
Here AD ⊥ BC, with AD = 15 cm, AC = 25 cm and BC = 28 cm.
In right triangle ADC, by the Pythagoras theorem,
⇒ DC2 = AC2 − AD2
⇒ DC2 = 252 − 152
⇒ DC2 = 625 − 225 = 400
⇒ DC = $\sqrt{400}$ = 20
∴ BD = BC − DC = 28 − 20 = 8 cm.
In right triangle ADB, by the Pythagoras theorem,
⇒ AB2 = AD2 + BD2
⇒ AB2 = 152 + 82
⇒ AB2 = 225 + 64 = 289
⇒ AB = $\sqrt{289}$ = 17
Hence, AB = 17 cm.
In the adjoining figure, AB and CD are two vertical poles of height 19 m and 11 m respectively. If the shortest distance between their tops is 17 m, find how far apart they are?
Let AB = 19 m and CD = 11 m be the two poles, with the tops 17 m apart. Through C, draw a horizontal line meeting AB at E.
Then AE = AB − CD = 19 − 11 = 8 m, and EC is the horizontal distance between the poles.
In right triangle AEC, by the Pythagoras theorem,
⇒ EC2 = AC2 − AE2
⇒ EC2 = 172 − 82
⇒ EC2 = 289 − 64 = 225
⇒ EC = $\sqrt{225}$ = 15
Hence, the poles are 15 m apart.
In the adjoining figure, find the value of x.
The lines AC and DF cross at P above the base.
In ΔDEF, ∠FED = 100° and ∠DFE = 46°, so
⇒ ∠FDC = 180° − (100° + 46°) = 34°.
In ΔABC, ∠A = 35° and ∠B = 115°, so
⇒ ∠ACB = 180° − (35° + 115°) = 30°.
In ΔPDC, by the angle sum property,
⇒ ∠DPC + ∠PDC + ∠PCD = 180°
⇒ ∠DPC + 34° + 30° = 180°
⇒ ∠DPC = 180° − (34° + 30°) = 116°
As, ∠DPC = x (vertically opposite angles)
⇒ x = 116°
Hence, x = 116°.
In the adjoining figure, AB = AC, BC = DC and DE ∥ BC. If ∠FAC = 130°, find ∠CDE.
∠FAC is the exterior angle of ΔABC at A, so
⇒ ∠BAC = 180° − 130° = 50°.
Since AB = AC, the base angles are equal:
⇒ ∠ABC = ∠ACB
In ΔABC, by angle sum property,
⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ ∠ABC + ∠ABC + 50° = 180° (As ∠ACB = ∠ABC)
⇒ 2∠ABC = 180° − 50° = 130°
⇒ ∠ABC = $\frac{130°}{2}$ = 65°
In ΔBDC, BC = DC and D lies on AB, so the base angles are equal:
⇒ ∠DBC = ∠BDC = 65°.
∴ ∠BCD = 180° − (65° + 65°) = 50°.
Since DE ∥ BC, ∠CDE = ∠BCD = 50° (alternate angles).
Hence, ∠CDE = 50°.
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