Question

The diagonals AC and BD of rhombus ABCD are 24 cm and 10 cm respectively. Find the area
of the rhombus and also its each side.
​

Answer 1

Given:

• The diagonals AC and BD of rhombus ABCD are 24 cm and 10 cm respectively.

To Find:

• The area of the rhombus and also its each side.

Diagram:

• Attached in photo

SOLUTION:

$\sf \: Area \: of \: the \: rhombus \: ABCD = \dfrac{1}{2} (Product \: of \: diagonals) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: = \dfrac{1}{2} (AC × BD) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} (24 × 10) sq. cm = 120 sq. cm$

• Since, the diagonals of a rhombus bisect each other at right angles

$\sf \: ∴ \: \: \: \: \: \: \: \: \: \: \: \: \: AO \: \: \: \: \: \: = \frac{1}{2} \: \: \: \: AC = \frac{1}{2} × 24 cm = 12 cm \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: OD \: \: \: \: \: \: \: = \frac{1}{2} \: \: \: \: \: BD = \frac{1}{2} × 10 cm = 5 cm$

• In right angled ∆AOD, by Pythagoras theorem, we have:

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AD² = AO ^{2} +OD {}^{2} \\ \: \: \: \: \: \: \: \: \ \implies \: \: \: \: \:\: \: \: \: \sf \: \: \: \: \: \: \: AD² \: = \:122 +5 \: \: \: \: \: \: \: \\ \implies \: \: \: \: \: \: \: \: \: \sf \: AD {}^{2} = 144+25 \\ \: \: \: \: \: \: \: \: \: \: \implies \: \: \: \: \: \ \sf \: AD {}^{2} \: \: \ = \: \: \: \: \: 169 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \sf AD = \sqrt{169} \: cm \\ \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \boxed{ \pink { \sf{ \bf13 \: cm}}} }\: \: \: \: \: \: \: \:$

• Hence, area of the rhombus is 120 sq. cm and length of each side is 13 cm.

Answer 2

★ The area of the rhombus is 120 cm² and the measure of its side equals 13 cm.

Step-by-step explanation:

Analysis -

‎ ‎ ‎ ‎ ‎ ‎In the question, it has been stated that the diagonals of a rhombus ABCD are AC and BD whose measures are 24 cm and 10 cm, respectively. We've been asked to find its area as well as its measure of each of its side.

Solution -

‎ ‎ ‎ ‎ ‎ ‎The given information is quite sufficient to find the area of the rhombus because, the formula which we use to find the area of rhombus requires only two parameters, i.e., the lengths of diagonal 1 and 2, and we have their lengths.

On plugging in their measures,

$\\ \\ \: \: \: \: \: { \sf \pmb{Area_{(rhombus)} = \dfrac{1}{2} \times d_{1} \times d_{2} \: sq.units}} \\ \\ \\ \twoheadrightarrow{ \sf{ \dfrac{1}{2} \times AC \times BD}} \\ \\ \\ \twoheadrightarrow{ \sf{ \dfrac{1}{2} \times 24 \times 10 }} \\ \\ \\ \twoheadrightarrow{ \sf{ \dfrac{1}{2} \times 240 }} \\ \\ \\ \twoheadrightarrow{ \sf{ \dfrac{ \cancel{240}}{ \cancel2} }} \\ \\ \\ \twoheadrightarrow{ \sf{ \frac{120}{1} }} \\ \\ \\ \twoheadrightarrow \underline{ \boxed{ \frak {\pmb{ \red{120 \: {cm}^{2} }}}}}$

Since we have obtained the area, let's head into the second sub-question, i.e., finding the measure of each side of the rhombus.

Make an imaginary figure of rhomus in your mind or refer the attachment. Now, label the figure as ABCD and join the diagonals. Let the intersecting point be O. You will notice that its diagonals make intersections at the right angles, which divides the rhombus into four right-angled triangles: ∆AOB, ∆BOC, ∆COD and ∆DOA.

The hypotenuse of any right-angled triangle equals the measure of side of the rhombus, the base of any right-angled triangle equals half the length of second diagonal and the perpendicular of any right-angled triangle equals the half the length of first diagonal.

Take ∆AOB into consideration and find its hypotenuse using Pythagorean Theorem.

$\:$

$\:$

• AO = ½ × 24 = 12 cm.
• OB = ½ × 10 = 5 cm.
• BA = Hypotenuse (?).

$\\ \\ \: \: \: \sf \pmb{ {(AO)}^{2} + {(OB)}^{2} = {(BA)}^{2} } \\ \\ \\ \twoheadrightarrow{ \sf{{(12)}^{2} + {(5)}^{2} = {(BA)}^{2} }} \\ \\ \\ \twoheadrightarrow \sf{{144 + 25 = {(BA)}^{2} }} \\ \\ \\ \twoheadrightarrow \sf{{169 = {(BA)}^{2} }} \\ \\ \\ \twoheadrightarrow \sf{{ \sqrt{169} = BA}} \\ \\ \\ \twoheadrightarrow \underline{ \boxed{ \frak{ \pmb{ \red{{13 = \tt BA}}}}}}$

• Therefore, the area and the length of each side of the rhombus are 120 cm² and 13 cm, respectively.

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