This document contains solutions to three triangle similarity problems: 1) It finds the ratio of the areas of two triangles with bases of 9 and 10 units and heights of 5 and 6 units is 3:4. 2) It finds the ratio of two triangles with the same base AB and heights of 4 and 8 units is 1:2. 3) It uses the property that the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corresponding heights to find that if RQ=6, PS=6, and PR=12, then QT=3 units.

1. 1.SIMILARITY

2. • Base of a triangle is 9 height is 5. base of another triangle is 10 and
height is 6. find the ratio of the area of these triangles.
PRACTICE SET 1.1
SOLUTION
LET THE BASE , HEIGHT AND AREA OF THE FIRST
TRIANGLE BE b1, h1, and A2 respectively.
Let the base, height and area of the
second triangle be b2, h2 and A2
respectively𝐴1
𝐴2
=
9𝑋5
10𝑋5
=
45
60
𝐴1
𝐴2
=
3
4
THE RATIO OF AREAS OF THE TRIANGLES IS 3:4

3. 2. IN THE ADJOINING FIGURE, BC ⊥ 𝐴𝐵 , 𝐴𝐷 ⊥ 𝐴𝐵 , BC = 4 ,
AD = 8 THEN FIND
∆ ∆𝐴𝐵𝐶
∆ ∆𝐴𝐷𝐵
SOLUTION:
∆𝐴𝐵𝐶 𝐴𝑁𝐷 ∆𝐴𝐷𝐵 𝐻𝐴𝑉𝐸 𝑇𝐻𝐸 𝑆𝐴𝑀𝐸 𝐵𝐴𝑆𝐸 𝐴𝐵
∆ ∆𝐴𝐵𝐶
∆ ∆𝐴𝐷𝐵
=
𝐵𝐶
𝐴𝐷
=
4
8
∆ ∆𝐴𝐵𝐶
∆ ∆𝐴𝐷𝐵
=
1
2
[ TRIANGLES HAVING EQUAL BASE ]
A
C
B
D

4. 3. IN THE ADJOINING FIGURE, SEG PS⊥ 𝑆𝐸𝐺 𝑅𝑄 , 𝑆𝐸𝐺 𝑄𝑇 ⊥ 𝑆𝐸𝐺 𝑃𝑅. 𝐼𝐹 𝑅𝑄 = 6, 𝑃𝑆 = 6 𝐴𝑁𝐷
PR =12 THE FIND QT.
P
R SQ
T
SOLUTION:
IN
∆𝑃𝑄𝑅, 𝑃𝑅 𝐼𝑆 𝑇𝐻𝐸 𝐵𝐴𝑆𝐸 𝐴𝑁𝐷 𝑄𝑇 𝐼𝑆 𝑇𝐻𝐸 𝐶𝑂𝑅𝑅𝐸𝑆𝑃𝑂𝑁𝐷𝐼𝑁𝐺 𝐻𝐸𝐼𝐺𝐻𝑇
ALSO RQ IS THE BASE AND PS IS THE CORRESPONDING
HEIGHT
𝐴(∆𝑃𝑄𝑅)
𝐴 ∆𝑃𝑄𝑅
=
𝑃𝑅𝑥𝑄𝑇
𝑅𝑄𝑥𝑃𝑆
[RATIOS OF THE AREAS OF THE TWO
TRIANGLES IS EQUAL TO THE RATIO
OF THE PRODUCT OF THEIR BASES
AND CORRESPONDING HEIGHTS]
1
1
=
𝑃𝑅𝑥𝑄𝑇
𝑅𝑄𝑥𝑃𝑆
PR x QT = RQ x PS
12 x QT = 6 x 6
QT =
36
12
QT = 3
UNITS