Use the following information to answer the next question.
ΔABC and ΔXYZ are two isosceles triangles such that AB = AC, XY = XZ, and ∠A = ∠X. The areas of ΔABC and ΔXYZ are 14 cm2 and 56 cm2 respectively.

If AD = 2.8 cm, then what is the height of ΔXYZ?
Answers
Answered by
1
Answer:
Step-by-step explanation: obviously the height is 5.6 cm
Answered by
2
Step-by-step explanation:
Area of Isosceles triangle = 1/2×base×height
in ∆ABC
Given :
Area = 14cm²
Height (AD) = 2.8cm
Area of ∆ABC = 1/2×base×height
14cm² = 1/2×base×2.8cm
base = (14×2)/2.8
base = 28/2.8
base = 10cm. -(i)
in ∆XYZ
Given :
Area = 56cm²
Base = 10cm (from i)
Area of ∆XYZ = 1/2×base×height
56cm² = 1/2×10×height
height = (56×2)/10
height = 112/10 = 11.2cm
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