Question

The length of the base of an isosceles triangle is 10 cm. If the length of the altitude is 1 cm shorter than the equal sides, find the area of the triangle.

Answer 1

Given :- The length of the base of an isosceles triangle is 10 cm. If the length of the altitude is 1 cm shorter than the equal sides, find the area of the triangle. ?

Solution :-

we know that,

• Median of an isosceles ∆ is perpendicular to the base.

so,

→ CD = DB = 5 cm. (Median divides the base in two equal parts.)

and,

→ ∠ADC = 90° .

now, in ∆ADC, we have,

→ AD² + CD² = AC² (by pythagoras theorem.)

Let us assume that, Length of altitude AD is x cm.

then,

→ AC = (x + 1) cm.

putting values now , we get,

→ x² + 5² = (x + 1)²

→ x² + 5² = x² + 2x + 1

→ 25 = 2x + 1

→ 25 - 1 = 2x

→ 2x = 24

→ x = 12 cm.

therefore,

→ Area of the ∆ABC = (1/2) * Base * Altitude = (1/2) * 10 * 12 = 60 cm². (Ans.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

https://brainly.in/question/16655884

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

The length of the base of an isosceles triangle is 10 cm. If the length of the altitude is 1 cm shorter than the equal sides, find the area of the triangle.

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \red{AηsωeR } ✍$

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$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{base \: of \: isosceles \: triangle \: = \: 10 \: cm} \\ &\sf{altitude \: is \: 1 \: shorter \: than \: equal \: side} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf To \: find - \begin{cases} &\sf{area \: of \: triangle} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf Formula- \begin{cases} &\sf{{Area\:of\:triangle= \frac{1}{2} \times base \times height}} \\ &\sf{ {(hyp.)}^{2} = {(base)}^{2} + {(perpendicular)}^{2} } \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\large\underline\red{\bold{❥︎Step :- 1 }}$

☆ Let us assume that ABC is an isosceles triangle with base BC and equal sides be AB and AC.

☆ The base of an isosceles triangle = BC = 10 cm

☆ Let the equal sides AB and AC of isosceles triangle be x cm.

$\sf \:  ⟼ \: ∴ AB \: = \: AC \: = \: x \: cm$

$\sf \:  ⟼ Now, \: altitude \: of \: triangle \: AD \: = \: (x - 1) \: cm$

☆ We know that, in an isosceles triangle altitude from the opposite vertex will always bisect the base.

$\sf \:  ⟼ \: So, BD \: = \: DC \: = \: \dfrac{1}{2} \times BC = \: 5 \: cm$

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$\large\underline\red{\bold{❥︎Step :- 2}}$

$\sf \:  ⟼Now, \: In \: Δ \: ABD$

$\sf \:  ⟼ \: by \: pythagoras \: theorem,$

$\sf \:  ⟼ \: {AB}^{2} \:  = {BD}^{2} \: \: + \: {AD}^{2}$

$\sf \:  ⟼ {x}^{2} = {(x - 1)}^{2} + {5}^{2}$

$\sf \:  ⟼ \: {x}^{2} = {x}^{2} + 1 - 2x + 25$

$\sf \:  ⟼ \: 0 = 26 - 2x$

$\sf \:  ⟼ \: 2x = 26$

$\bf\implies \: \: x \: = \: 13 \: cm$

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$\large\underline\red{\bold{❥︎Step :- 3 }}$

$\bf \:  ⟼ \: So \: dimensions \: are \:$

$\sf \:  ⟼ \: AB \: = \: x \: = \: 13 \: cm$

$\sf \:  ⟼ \: AD \: = \: x \: - 1 = \: 13 - 1 \: = 12 \: cm$

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$\large\underline\red{\bold{❥︎Step :- 4 }}$

$\bf \:  ⟼ \: area \: of \triangle \: ABC = \: \dfrac{1}{2} \times AD \times BC$

$\bf \:  ⟼ \: area \: of \triangle \: ABC = \: \dfrac{1}{2} \times 10 \times 12$

$\bf \:  ⟼ \: area \: of \triangle \: ABC = \:5 \times 12 = 60 {cm}^{2}$

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$\large \red{\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Explore \: more } ✍$

${\large{\bold{\rm{\underline{Additional \; knowledge}}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cuboid \: = \: 2(l \times b + b \times h + l \times h}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cuboid \: = \: 2h(l+b)}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cuboid \: = \: L \times B \times H}}} \\ </p><p>\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cube\: = \: \sqrt 3l}}}$