1) In triangle ADE, BC is parallel to DE Ar of triangle ABC = 25 sq cm, ar of - Maths - Introduction to Three Dimensional Geometry

Question

1) In triangle ADE, BC is parallel to DE Ar of triangle ABC = 25
sq cm, ar of trapezium BCED = 24 sq cm and DE = 14 cm Calculate the
length of BC (this I found to be 10 cm) Also find the area of
triangle BCD ( I am stuck here)
2) In a trapezium ABCD, AB is parallel to DC and diagonals AC
and BD intersect at point P If AP : CP = 3 : 5, find : a) triangle
APB : triangle CPB (b) triangle DPC : triangle APB (c) triangle ADP
: triangle APB (d) triangle APB : triangle ADB

Santosh Kumar · Accepted Answer

area of Î”ADE = ar (Î”ABC) + ar (trapezium BCED) = 25 + 24 = 49 cm2 In Î”ABC and Î”ADE, $∠ A = ∠ A (common) ∠ ABC = ∠ ADE (corresponding angles) ∠ ACB = ∠ AED (corresponding angles) ∴ Δ ABC \sim Δ ADE \frac{a r (Δ ABC)}{a r (Δ ADE)} = \frac{{BC}^{2}}{{DE}^{2}} = \frac{{BC}^{2}}{(14)^{2}} = \frac{{BC}^{2}}{196} \Rightarrow \frac{25}{49} = \frac{{BC}^{2}}{196} \Rightarrow {BC}^{2} = \frac{196 \times 25}{49} \Rightarrow BC = 10 cm$ let CF be the perpendicular distance between BC and DE $a r (trapezium BCED) = \frac{1}{2} (10 + 14) \times CF \Rightarrow \frac{1}{2} \times 24 \times CF = 24 \Rightarrow CF = 2 cm a r (Δ BCD) = \frac{1}{2} \times BC \times CF (CF is the perpendicular to the base BC of the triangle BCD) = \frac{1}{2} \times 10 \times 2 = 10 {cm}^{2}$