Question

8. An Altitude of a triangle is five-thirds the length of its corresponding base. If the altitude were
Increased by 8 cm and the base decrcased by 4 cm the area of the triangle would remain the same,
Find the base and altitude of the triangle.

answer this​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

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$\green{\underline \bold{Given :}}$

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• Altitude is five third the length of base.

• Area remains same when altitude is increased by 8 cm & base is decreased by 4 cm.

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$\red{\underline \bold{To \: Find:}}$

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• Base and altitude of the triangle.

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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Let the altitude be 'x'

Let the base be 'y'

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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$\purple\longrightarrow$ $\sf x = \dfrac { 5 } { 3 } \times y$ ------(1)

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$\underline{\bold{\texttt{Area of triangle :}}}$

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$\bf \dashrightarrow \dfrac { 1 } { 2 } \times base \times height$

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$\underline{\bold{\texttt{Area of original triangle ,}}}$

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$\sf\longmapsto \dfrac { 1 } { 2 } \times y\times x$ -----(2)

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$\underline{\bold{\texttt{Putting value of x in (2)}}}$

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$\sf\longmapsto \dfrac { 1 } { 2 } \times y\times \dfrac { 5y } { 3 }$ ----[Area 1 ]

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$\underline{\bold{\texttt{New base :}}}$

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$\purple\longrightarrow$ $\sf y - 4$

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$\underline{\bold{\texttt{New height :}}}$

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$\purple\longrightarrow$ $\sf x + 8$

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$\underline{\bold{\texttt{Area of new triangle :}}}$

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$\sf\longmapsto \dfrac { 1 } { 2 } \times y - 4\times x + 8$-----(3)

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$\underline{\bold{\texttt{Putting the value of x in (3)}}}$

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$\sf\longmapsto \dfrac { 1 } { 2 } \times y - 4\times \dfrac { 5y } { 3 } + 8$ -----[Area 2]

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$\red{\underline \bold{According \: to \: the \ question :}}$

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Area remains same

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$\red{\bold{</u><u>Area</u><u> </u><u>\: </u><u>1</u><u> </u><u>\: </u><u>=</u><u> </u><u>\: </u><u>Area</u><u> </u><u>\: </u><u>2</u><u>}}$

$\:\:$

$\sf\longmapsto \dfrac { 1 } { 2 } \times y\times \dfrac { 5y } { 3 }$ $\sf =$ $\sf </p><p>\dfrac { 1 } { 2 } \times y - 4\times \dfrac { 5y } { 3 } + 8$

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$\sf \longmapsto \dfrac { 5y^2 } { 3 } = (y - 4) \times \dfrac { 5y + 24 } { 3 }$

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$\sf \longmapsto 5y^2 = y(5y + 24) -4(5y + 24)$

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$\sf \longmapsto 5y^2 = 5y^2 + 24y - 20y - 96$

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$\sf \longmapsto 4y = 96$

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$\sf \longmapsto y = \dfrac { 96 } { 4 }$

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$\bf \dashrightarrow y = 24 cm$

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$\underline{\bold{\texttt{Putting y = 24 in (1)}}}$

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$\sf \longmapsto x = \dfrac { 5 } { 3 } \times 24$

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$\sf \longmapsto x = 5 \times 8$

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$\bf\dashrightarrow x = 40 cm$

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Hence base is 24 cm & altitude is 40 cm

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$\rule{200}5$

Answer 2

Answer:

base 24

Step-by-step explanation:

altitude 40

pls make it brainlest answer