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Solution for The Perimeter of a Triangular Field is 240 Dm. If Two of Its Sides Are 78 Dm and 50 Dm, Find the Length of the Perpendicular on the Side of Length 50 Dm from the Opposite Vertex. - CBSE Class 9 - Mathematics

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Question

The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.

Solution

ABC be the triangle, Here a = 78 dm = AB,
BC = b = 50 dm

Now, perimeter = 240 dm
⇒ AB + BC + CA = 240 dm
⇒ AC = 240 – BC – AB
⇒ AC = 112 dm
Now, 2s = AB + BC + CA
⇒ 2s = 240
⇒ s = 120 dm

∴ Area of ΔABC =`sqrt(s(s-a)(s-b)(s-b))` by heron's formula 

=`sqrt(120(120-78)(120-50)(120-112))`

`=sqrt(120xx42xx70xx8)`

1680 `dm^2`

𝐿𝑒𝑑 𝐴𝐷 𝑏𝑒 π‘π‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘–π‘π‘’π‘Žπ‘™π‘Žπ‘Ÿ π‘œπ‘› 𝐡𝐢

Area of ΔABC = `1/2xx ABxxBC `(area of triangle=`(1/2xxbxxh)`

`=1/2xx ADxxBC=1680 `

`⇒AD=(2xx1680)/50=67.2dm`

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Solution The Perimeter of a Triangular Field is 240 Dm. If Two of Its Sides Are 78 Dm and 50 Dm, Find the Length of the Perpendicular on the Side of Length 50 Dm from the Opposite Vertex. Concept: Area of a Triangle by Heron’S Formula.
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