A parallelogram the measures of whose afjacent sides are 28cm and 42 cm has on diagonals 38 cm . Find its altitude on the side 42cm . l amd m , two parallel lines are intersected by another pair of paralkel lines p and c . Show that triangle ABC = triangle
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Answers
Answer:
Given,
One side ( AB ) = 42 cm
Another ( BC ) = 28 cm
One Diagonal ( AC ) = 38 cm.
We know that diagonal of a parallelogram divides it into two equal parts.
So, Ar. of Parallelogram ABCD = Ar. of ΔABC × 2
For ΔABC,
AB(c) = 42 CM , BC(a) = 28 CM and AC(b) = 38 CM.
Semi perimeter ( s ) = ( a + b + c ) / 2
= ( 42 cm + 28 cm + 38 cm ) / 2
= ( 108 cm ) / 2
= 54 cm.
Now,
⇒ ( s - a ) = 54 cm - 28 cm = 26 cm
⇒ ( s - b ) = 54 cm - 38 cm = 16 cm
⇒ ( s - c ) = 54 cm - 42 cm = 12 cm.
By Heron's Formula ,
Ar. of ΔABC = √{ s ( s - a ) ( s - b ) ( s -c ) }
= √ ( 54 cm × 12 cm × 16 cm × 26 cm )
= √ ( 2 × 3³ × 2² × 3 × 2⁴ × 2 × 13 cm⁴ )
= √ ( 3⁴ × 2⁸ × 13 cm⁴ )
= 3² × 2⁴ cm² √13
= 144√13 cm²
Now,
⇒ Ar. of parallelogram ABCD = 2 × Ar. of Δ ABC
= 2 × 144√13 cm²
= 288√13 cm².
⇒ Ar. of a Parallelogram = Base × Corresponding altitude
⇒ 288√13 cm² = 42 cm × DE
⇒ DE = 288√13 cm² ÷ 42
⇒ DE = 48√13 cm² / 7
∴ DE = 48√13 cm² / 7.
hope it helps