Solutions: Triangles - Vardaan Learning Institute

1. If $\triangle ABC \sim \triangle PQR$, area($\triangle ABC$) = 81 $cm^2$ and area($\triangle PQR$) = 144 $cm^2$. If AB = 9 cm, then PQ is:

Ratio of areas = $(\frac{AB}{PQ})^2$
$\frac{81}{144} = (\frac{9}{PQ})^2$
Taking square root: $\frac{9}{12} = \frac{9}{PQ}$
$PQ = 12$ cm.

Correct Option: (a) 12 cm

2. In $\triangle ABC$, DE || BC. If AD = 1.5 cm, DB = 3 cm and AE = 1 cm, then EC is:

By BPT, $\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{1.5}{3} = \frac{1}{EC}$
$\frac{1}{2} = \frac{1}{EC} \Rightarrow EC = 2$ cm.

Correct Option: (c) 2 cm

3. The lengths of the diagonals of a rhombus are 24 cm and 32 cm. The side of the rhombus is:

Diagonals bisect at 90 deg. Half diagonals are 12 cm and 16 cm.
Side $s = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$ cm.

Correct Option: (c) 20 cm

4. Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, the distance between their tops is:

Difference in height = $11 - 6 = 5$ m.
Horizontal distance = 12 m.
Distance between tops = $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ m.

Correct Option: (a) 13 m

5. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. The height of the tower is:

$\triangle$s are similar. Ratio of height to shadow is constant.
$\frac{6}{4} = \frac{H}{28} \Rightarrow H = \frac{6 \times 28}{4} = 6 \times 7 = 42$ m.

Correct Option: (b) 42 m

6. Two isosceles triangles have equal angles and their areas are in the ratio 16:25. The ratio of their corresponding heights is:

Ratio of heights = $\sqrt{\text{Ratio of Areas}}$
Ratio = $\sqrt{16/25} = 4/5$.

Correct Option: (a) 4:5

7. If $\triangle ABC \sim \triangle DEF$, BC = 3 cm, EF = 4 cm and area of $\triangle ABC = 54 cm^2$. Then area of $\triangle DEF$ is:

Ratio of areas = Ratio of squares of corresponding sides.
$\frac{Area(ABC)}{Area(DEF)} = (\frac{BC}{EF})^2$
$\frac{54}{Area(DEF)} = (\frac{3}{4})^2 = \frac{9}{16}$
$Area(DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96 cm^2$.

Correct Option: (a) 96 $cm^2$

8. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Using Pythagoras theorem: $Hypotenuse^2 = Base^2 + Height^2$
$10^2 = x^2 + 8^2$
$100 = x^2 + 64 \Rightarrow x^2 = 36 \Rightarrow x = 6$ m.

Correct Option: (a) 6 m

9. In $\triangle ABC$, if $AB^2 = AC^2 + BC^2$, then the angle opposite to side AB is:

This is the converse of Pythagoras Theorem. The angle opposite the longest side (hypotenuse AB) is a right angle ($90^\circ$).

Correct Option: (c) 90 deg

10. Assertion (A): Two similar triangles are always congruent. Reason (R): If the areas of two similar triangles are equal, then they are congruent.

Assertion is False: Similarity implies same shape, not necessarily same size.
Reason is True: If areas are equal, scaling factor is 1, so they are congruent.

Correct Option: (d) A is false but R is true.

11. In $\triangle ABC$, if angles P, Q, R are mid-points of the sides BC, CA and AB respectively. Find the ratio of the area of $\triangle PQR$ to area of $\triangle ABC$.

Line joining midpoints of two sides is parallel to third side and half of it.
So, $\triangle PQR$ is formed by midpoints, dividing $\triangle ABC$ into 4 congruent triangles.
Area($\triangle PQR$) = $\frac{1}{4}$ Area($\triangle ABC$).

Ratio = 1:4

12. In a trapezium ABCD, AB || DC and its diagonals intersect each other at the point O. Show that $AO/BO = CO/DO$.

In $\triangle AOB$ and $\triangle COD$:
$\angle OAB = \angle OCD$ (Alt. int. angles)
$\angle OBA = \angle ODC$ (Alt. int. angles)
So $\triangle AOB \sim \triangle COD$ (AA similarity).
Thus, corresponding sides are proportional: $\frac{AO}{CO} = \frac{BO}{DO}$.
Rearranging: $\frac{AO}{BO} = \frac{CO}{DO}$.

Proved

13. The perimeters of two similar triangles ABC and PQR are 60 cm and 48 cm respectively. If PQ = 8 cm, find AB.

Ratio of perimeters = Ratio of corresponding sides.
$\frac{Perimeter(ABC)}{Perimeter(PQR)} = \frac{AB}{PQ}$
$\frac{60}{48} = \frac{AB}{8}$
$AB = \frac{60 \times 8}{48} = \frac{480}{48} = 10$.

AB = 10 cm

14. If $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.

Given $\triangle ABE \cong \triangle ACD$, imply $AB = AC$ and $AE = AD$ (CPCT).
So, $\frac{AD}{AB} = \frac{AE}{AC}$.
Also, $\angle A$ is common.
By SAS similarity criterion, $\triangle ADE \sim \triangle ABC$.

Proved

15. D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

In $\triangle ABC$ and $\triangle DAC$:
1. $\angle BAC = \angle ADC$ (Given)
2. $\angle C = \angle C$ (Common)
$\therefore \triangle ABC \sim \triangle DAC$ (AA Similarity)
$\Rightarrow \frac{CA}{CD} = \frac{CB}{CA}$ (Corresp. sides)
$\Rightarrow CA^2 = CB \cdot CD$.

Proved

16. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle PQR$. Show that $\triangle ABC \sim \triangle PQR$.

Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$. Since AD, PM are medians, $BC = 2BD, QR = 2QM$.
$\Rightarrow \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
So $\triangle ABD \sim \triangle PQM$ (SSS). Thus $\angle B = \angle Q$.
In $\triangle ABC, \triangle PQR$: $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
$\therefore \triangle ABC \sim \triangle PQR$ (SAS).

Proved

17. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Let $\triangle ABC \sim \triangle PQR$. Let AD and PM be medians.
As proved in Q16, $\triangle ABD \sim \triangle PQM$. So $\frac{AB}{PQ} = \frac{AD}{PM}$.
We know Ratio of Areas = $(\frac{AB}{PQ})^2$.
Substituting, Area Ratio = $(\frac{AD}{PM})^2$.

Proved

18. BL and CM are medians of a triangle ABC right angled at A. Prove that $4(BL^2 + CM^2) = 5 BC^2$.

In rt $\triangle BAL$: $BL^2 = AB^2 + AL^2 = AB^2 + (\frac{AC}{2})^2 = AB^2 + \frac{AC^2}{4} \Rightarrow 4BL^2 = 4AB^2 + AC^2$.
In rt $\triangle CAM$: $CM^2 = AC^2 + AM^2 = AC^2 + (\frac{AB}{2})^2 = AC^2 + \frac{AB^2}{4} \Rightarrow 4CM^2 = 4AC^2 + AB^2$.
Adding: $4(BL^2 + CM^2) = 5(AB^2 + AC^2)$. Since $AB^2+AC^2 = BC^2$,
$4(BL^2 + CM^2) = 5 BC^2$.

Proved

19. Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that $EL = 2 BL$.

In $\Delta BMC$ and $\Delta EMD$:
$\angle BMC = \angle EMD$ (Vert. opp)
$MC = MD$ (M is mid-point)
$\angle BCM = \angle EDM$ (Alt int angles, $BC || AE$)
$\therefore \Delta BMC \cong \Delta EMD$ (ASA).
$\Rightarrow BC = DE$. Since $AD = BC$ (Opp sides of ||gm), $AE = AD + DE = 2BC$.
Now, $\Delta AEL \sim \Delta CBL$ (AA similarity).
$\frac{EL}{BL} = \frac{AE}{BC} = \frac{2BC}{BC} = 2$.
$\therefore EL = 2 BL$.

Proved

20. A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time a tower casts the shadow 40 m long on the ground. Find the height of the tower.

At the same time, sun's elevation is same, so triangles are similar.
$\frac{Height_{stick}}{Shadow_{stick}} = \frac{Height_{tower}}{Shadow_{tower}}$
$\frac{12}{8} = \frac{H}{40}$
$H = \frac{12 \times 40}{8} = 12 \times 5 = 60$ m.

Height = 60 m

21. State and prove the Basic Proportionality Theorem (Thales Theorem).

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: $\triangle ABC$, $DE || BC$.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$.
Proof: Area($\triangle ADE$) = $\frac{1}{2} \times AD \times EN$.
Area($\triangle BDE$) = $\frac{1}{2} \times DB \times EN$.
Ratio 1: $\frac{Area(ADE)}{Area(BDE)} = \frac{AD}{DB}$.
Similarly, Ratio 2: $\frac{Area(ADE)}{Area(DEC)} = \frac{AE}{EC}$.
Since $\triangle BDE$ and $\triangle DEC$ are on same base DE and between same parallels DE and BC, Area(BDE) = Area(DEC).
From (1) and (2), $\frac{AD}{DB} = \frac{AE}{EC}$.

Proved

22. State and prove the Pythagoras Theorem.

Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Given: Rt $\triangle ABC$, $\angle B = 90^\circ$.
To Prove: $AC^2 = AB^2 + BC^2$.
Construction: Draw $BD \perp AC$.
Proof: $\triangle ADB \sim \triangle ABC \Rightarrow \frac{AD}{AB} = \frac{AB}{AC} \Rightarrow AB^2 = AD \cdot AC$ (1)
$\triangle BDC \sim \triangle ABC \Rightarrow \frac{CD}{BC} = \frac{BC}{AC} \Rightarrow BC^2 = CD \cdot AC$ (2)
Adding (1) and (2): $AB^2 + BC^2 = AC(AD + CD) = AC \cdot AC = AC^2$.

Proved

23. Case Study: Light House. (i) Sketch (ii) Length of shadow (iii) Similarity Criterion

(i) Draw a right triangle representing the lamp post and boy.
(ii) Let lamp height $H = 3.6$ m, boy height $h = 0.9$ m. Speed = 1.2 m/s. Time = 4s.
Distance derived by boy $D = speed \times time = 1.2 \times 4 = 4.8$ m.
Let shadow length be $x$.
By similarity: $\frac{H}{D+x} = \frac{h}{x} \Rightarrow \frac{3.6}{4.8+x} = \frac{0.9}{x}$
$4 = \frac{4.8+x}{x} \Rightarrow 4x = 4.8 + x \Rightarrow 3x = 4.8 \Rightarrow x = 1.6$ m.
(iii) AA Similarity is used (90 deg angles and common angle).

(ii) 1.6 m (iii) AA Similarity