Question

$\bf\huge{Question:}$ The area of a rhombus is equal to that of a triangle whose side and corresponding altitude are 24cm and 15cm respectively if one of the diagonals of the rhombus is 18cm . Find the other

Answer 1

Answer:

Area of rhombus=area of triangle

(1/2)×d1×d2=(1/2)×base×altitude

(1/2)×22×d2=(1/2)×24.8×16.5

d2=24.8×16.5/22 cm.

d2=(124/5)×(33/2)×(1/22)

d2=(124×33×1)/(5×2×22)

d2=372/20=18.6 cm. , .

Answer 2

★ Given:-

• The area of a rhombus is equal to that of a triangle side and corresponding altitude are 24cm and 15cm respectively.

• one of the diagonals of the rhombus is 18cm

★ To Find : -

• The other diagonal of the rhombus

★ Concept : -

➤ Here, we have been provided with the hieght and the base of the triangle and the measurement of a diagonal in the rhombus respectively. It is said that the area of the rhombus is equal to the area of the triangle. Now, let's find the area of the triangle which will further help us in finding the diagonal of the rhombus.

★ Solution : -

➢ Now,

• Let's find the area of the triangle

➢ We know that,

${ \pink{\star}}{ \pink{ \underline{ \boxed{ \bf{Area_{(triangle)} = \frac{1}{2} \times base \times hieght}}}}}$

➢ Here,

• Base ( B ) = 24cm
• Hieght ( H ) = 15cm

➤ Substituting we get,

${ : \implies} \rm \: Area _{(triangle)} = \frac{1}{2} \times b \times h \: \: \: \: \: \\ \\ \\ { : \implies} \rm \: Area _{(triangle)} = \frac{1}{ \cancel2} \times \cancel{24} \times 15 \\ \\ \\ { : \implies} \rm \: Area _{(triangle)} = 12 \times 15 \: \: \: \: \: \: \: \: \: \\ \\ \\{ : \implies} \rm \: Area _{(triangle)} = 180 {cm}^{2} \: \: \: \: \: \: \: \: \:$

• Henceforth area of the triangle is 180cm²

➢ We know that,

➛Area of triangle is equal to the area of rhombus

• Henceforth the area of the rhombus is 180cm²

➢ Now,

• Let's find its other diagonal

➢ As we know that,

${ \pink{\star}}{ \pink{ \underline{ \boxed{ \bf{Area_{(rhombus)} = \frac{1}{2} \times d_{1} \times \: d_{2} }}}}}$

➢Where,

• Diagonal ( 1 ) = 18cm

• Area = 180cm²

➤ Substituting we get,

${ : \implies} \rm \: Area _{(rhombus)} = \frac{1}{2} \times d _{1} \times d _{2} \\ \\ \\ { : \implies} \rm 180 = \frac{1}{2} \times 18 \times d _{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm \: d _{2} = \frac{180 \times \cancel2}{ \cancel{18}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm \: d _{2} = \cancel\frac{180}{9} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm \: { \pink{ \underline{ \boxed{ \frak{d _{2} =20cm}} \star}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Henceforth, the other diagonal is 20cm

★ More to know :-

${\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}} \\ \\ {\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}} \\ \\ {\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}} \\ \\{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}} \\ \\{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}} \\ \\ \; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi r^{2}}}} \\ \\ \; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}} \\ \\ {\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}} \\ \\{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}} \\ \\ {\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}} \\ \\ {\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}} \\ \\ {\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\ \\ {\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}} \\ \\ {\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}} \\ \\ {\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}} \\ \\ {\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}} \\ \\{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$