∆ABC∼∆APQ, AP=4,AQ=5,AB=10, Find AC
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Answer:
Step-by-step explanation:
SOLUTION:
Given: ΔACB∼ΔAPQ , BC = 10 cm , PQ = 5 cm , BA = 6.5 cm , AP = 2.8 cm.
(i) ΔACB∼ΔAPQ (given)
We know that corresponding sides of similar triangles are proportional.
AB/AQ = CB/PQ = AC/AP
AB/AQ = CB/PQ
6.5/AQ = 10/5
6.5 × 5 = 10 AQ
AQ = (6.5×5)/10
AQ = 6.5/2
AQ = 3.25 cm
Similarly,
CB/PQ = AC/AP
CA/2.8 = 10/5
5×CA = 2.8 × 10
CA = (10 × 2.8)/5
CA = 2 × 2.8
CA = 5.6 cm
Hence, the length of AQ is 3.25 cm and CA is 5.6 cm.
(ii) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
arΔACQ/arΔAPQ = (BC/PQ)²
= (10/5)²
= (2/1)²
= 4/1
Hence, the Area of ΔACB : Area of ΔAPQ is 4:1.
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