Given :

• Base = 18 cm,
• Height = 11 cm

Formula: The area of a parallelogram is calculated as: Area = base × height,.

∴ Area = 18 cm × 11 cm = 198 sq cm.

The area of the parallelogram is 198 sq cm.

Given:

• Area = 29.6 sq cm
• Base = 8 cm
• Height = ?

Formula: The area of a parallelogram is given by: Area = base × height.

Substituting the values into the formula: 29.6 = 8 × height

∴ Height = \(\frac{29.6}{8}\) = 3.7 cm.

The height of the parallelogram is 3.7 cm.

Given :

• Area = 83.2 sq cm
• Height = 6.4 cm
• Base = ?

Formula: The area of a parallelogram is calculated as: Area = base × height.

Substituting the given values into the formula: 83.2 = base × 6.4

∴ Base = \(\frac{83.2}{6.4}\) = 13 cm.

The length of the base of the parallelogram is 13 cm.

Given :

• Length of the first diagonal (d₁) = 15 cm
• Length of the second diagonal (d₂) = 24 cm

Formula: The area of a rhombus is equal to half the product of the lengths of its diagonals.

∴ Area of a rhombus = \(\frac{1}{2}\) × d₁ × d₂

Substituting these values into the formula: Area = \(\frac{1}{2}\) × 15 × 24

Area = 15 × 12 = 180 sq cm.

The area of the rhombus is 180 sq cm.

Given :

• Length of the first diagonal (d₁) = 16.5 cm
• Length of the second diagonal (d₂) = 14.2 cm

Formula: The area of a rhombus is equal to half the product of the lengths of its diagonals.

∴ Area of a rhombus = \(\frac{1}{2}\) × d₁ × d₂

Substituting these values into the formula: Area = \(\frac{1}{2}\) × 16.5 × 14.2

Area = 16.5 × 7.1 = 117.15 sq cm.

The area of the rhombus is 117.15  sq cm.

Let □ ABCD be the required rhombus.

AD and BC are the two diagonals intersecting at point O.

AD = 48 cm

Perimeter of the rhombus = 100 cm = 4 x side

∴ 100 = 4 x l(AB)

∴ l(AB)  = 100/4 = 25 cm

Thus, each side of the rhombus = 25 cm.

Diagonals of a rhombus are perpendicular bisector of each other.

Length of one diagonal AD = 48 cm         ….(Given)

∴ AO = OD = 48/2 = 24 cm

In Δ AOB,

We apply Pythagoras theorem,

AO² + OB² = AB²

∴ 24² + OB² =25²

∴ OB² = 625 – 576 = 49

∴ OB = 7 cm

So, CB = 2 x OB = 2 x 7 = 14 cm

Area of rhombus = \(\frac{1}{2}\) × product of lengths of diagonals

= \(\frac{1}{2}\) × 48 x 14

= 336 sq. cm

The area of the rhombus is 336 sq cm.

Let □ ABCD be the required rhombus. AC and BD are the two diagonals intersecting at point M.

A(□ ABCD) =240 sq cm, BD = 30 cm

Area of rhombus = \(\frac{1}{2}\) × product of lengths of diagonals

240 = \(\frac{1}{2}\) × l(BD) × l(AC)

240 = \(\frac{1}{2}\) × 30 × l(AC)

240 = 15 × l(AC)

l(AC) = 240/15 = 16 cm

Diagonals of rhombus bisect each other

∴ l(AM) = \(\frac{1}{2}\) × l(AC) and l(BM) = \(\frac{1}{2}\) × l(BD)

∴ l(AM) = \(\frac{1}{2}\) × 16 and l(BM) = \(\frac{1}{2}\) × 30

∴ l(AM) = 8 cm and l(BM) = 15 cm

Diagonals of rhombus are perpendicular to each other

∴ m∠ AMB = 90°

In right angled Δ AMB,

by Pythagoras' theorem,

l(AB)² = l(AM)² + l(MB)²

∴ l(AB)² = 82 + 152

∴ l(AB)² = 64 + 225

∴ l(AB)² = 289

∴ l(AB) = 289

∴ l(AB) = 17 cm

∴ side of rhombus is 17 cm

Perimeter of rhombus = 4 x side = 4 × 17 = 68 cm

Perimeter of rhombus is 68 cm.

l(AB) = 13 cm, l(DC) = 9 cm and l(AD) = 8 cm, □ ABCD is a trapezium

Area of a trapezium = \(\frac{1}{2}\) × sum of the lengths of parallel sides × height

∴ A(□ABCD) = \(\frac{1}{2}\) × [l(AB) + l(DC)] × l(AD)

∴ A(□ABCD) = \(\frac{1}{2}\) × [13 + 9] × 8

∴ A(□ABCD) = 22 × 4 = 88 sq cm

The area of □ ABCD is 88 sq cm.

Given:

• Lengths of parallel sides = 8.5 cm and 11.5 cm.
• Height = 4.2 cm.

Formula: The area of a trapezium is calculated as: Area = \(\frac{1}{2}\) × (sum of the lengths of parallel sides) × height.

Substituting these values into the formula:

Area = \(\frac{1}{2}\) × (8.5 + 11.5) × 4.2

= \(\frac{1}{2}\) × (8.5 + 11.5) × 4.2

= 10 × 4.2

∴ Area = 42 sq cm.

The area of the trapezium is 42 sq cm.

□ PQRS is an isosceles trapezium

l(PQ) =7 cm, l(SM) = 3 cm and l(PS) = l(QR)

Draw QN ⊥ side SR

Distance between parallel lines is constant

∴ QN = 4 cm

In right angled Δ PMS and Δ QNR

Hypotenuse PS = hypotenuse QR

Side PM ≅ side QN

∴ Δ PMS ≅ Δ QNR                  ... (By hypotenuse side test)

∴ l(NR) = l(SM)                     ... (Corresponding sides of congruent triangles)

∴ l(NR) =3 cm

PMNQ is a rectangle

∴ l(MN) = l(PQ) = 7 cm         …..(Opposite sides of rectangle are equal)

l(SR) = l(SM) + l(MN) + l(NR)

∴ l(SR) = 3 + 7 + 3 ∴ l(SR) = 13 cm

The area of a trapezium = \(\frac{1}{2}\) × (sum of the lengths of parallel sides) × height.

∴ A(□ DPQRS) = \(\frac{1}{2}\) x[l(PQ) + l(SR)] x l(PM)

= \(\frac{1}{2}\) × [7 + 13] × 4 = 20 × 2 = 40 sq cm

Area of □ PQRS is 40 sq cm

Given : Sides: a = 45 cm, b = 39 cm, c = 42 cm.

Semiperimeter (s): s = \(\frac{a+b+c}{2}\)

= \(\frac{45+39+42}{2}\)

= \(\frac{126}{2}\)

∴ s = 63 cm

Using Heron's Formula:

Area of triangle: A = \(\sqrt{s(s-a)(s-b)(s-c)}\)

∴ Area = \(\sqrt{63(63-45)(63-39)(63-42)}\)

= \(\sqrt{63×18×24×21}\)

= \(\sqrt{9×7×9×2×8×3×7×3}\)

= \(\sqrt{9^2×7^2×4^2×3^2}\)

= 9 × 7 × 4 × 3

∴ Area = 756 sq cm.

The area of the triangle is 756 sq cm.

In triangle PSR,

Applying Pythagoras theorem,

PS² + SR² = PR²

36² + 15² = PR²

PR² = 1296 + 225

PR² = 1521

PR = \(\sqrt{1521}\)

PR = 39 m

Area of triangle PSR = \(\frac{1}{2}\) x 15 × 36 = 270 m²

In triangle PQR,

by using Heron's formula,

Let a = 56, b =25, c = 39

So,

s = \(\frac{a+b+c}{2}\) = \(\frac{56+25+39}{2}\) = \(\frac{120}{2}\)

∴ s = 60 cm

Area of triangle: A = \(\sqrt{s(s-a)(s-b)(s-c)}\)

∴ Area = \(\sqrt{60(60-56)(60-25)(60-39)}\)

= \(\sqrt{60×4×35×21}\)

= \(\sqrt{176400}\)

∴ Area = 420 m².

Area of PQRS = Area of triangle PSR + Area of triangle PQR

= 270 + 420

= 690 m²

Area of □ PQRS is = 690 m²

A(□ ABCD) = A(Δ BAD) + A(Δ BDC)

In Δ BAD, m∠ BAD = 90°, l(AB) = 40m, l(AD) = 9m

A(Δ BAD) = \(\frac{1}{2}\) × product of sides forming the right angle

= \(\frac{1}{2}\) × l(AB) × l(AD)

= \(\frac{1}{2}\) × 40 × 9

= 180 sq.m

In Δ BDC, l(BT) = 13m, l(CD) = 60m

A(ABDC) = \(\frac{1}{2}\) x base x height

= \(\frac{1}{2}\) × l(CD) × l(BT)

= \(\frac{1}{2}\) × 60 × 13

= 390 sq. m

A(□ ABCD) = A(Δ BAD) + A(Δ BDC)

= 180 + 390

= 570 sq. m

:. The area of □ ABCD is 570 sq. m.

In given Figure,

PA = 30 m, AB = 30 m BC = 30 m, CS = 60 m, QA =5 0m, RC = 25m, TB = 30m.

A(Δ PAQ) = \(\frac{1}{2}\) × PA × QA

= \(\frac{1}{2}\) × 30 × 50

∴ A(Δ PAQ) = 750 m²            .... (1)

A(Δ RCS) = \(\frac{1}{2}\) × CS × RC

= \(\frac{1}{2}\) × 60 × 25

∴ A(Δ RCS) = 750 m²            ...... (2)

∵ P-A-B-C-S

∴ PS = PA + AB + BC + CS

= 30 + 30+30+60

∴ PS = 150 m

Also, AC = AB + BC = 30 + 30 = 60 m

A(Δ PTS) = \(\frac{1}{2}\) × PS × TB

= \(\frac{1}{2}\) × 150 × 30 = 2250

∴ A(Δ PTS) = 2250 m²           .... (3)

QA || RC

QRCA is a Trapezium.

A(□ QRCA) = \(\frac{1}{2}\) × (QA + RC) × AC

= \(\frac{1}{2}\) × (50 + 25) × 60 = \(\frac{1}{2}\) × 75 × 60

A(□ QRCA) = 2250 m²          …..(4)

Area of given Plot = A(Δ PAQ) + A(□ QRCA) + A(Δ RCS) + A(Δ PTS)

= 750 + 2250 + 750 + 2250          …..[From (1), (2), (3) and (4)]

∴ Area of the given Plot = 6000 m²

In Δ ABE, m∠ BAE = 90°, l(AB) = 24 m, l(BE) = 30 m

∴ [l(BE)]² = [l(AB)]² + [l(AE)]²                ... [Pythagoras theorem]

∴ (30)² = (24)² + [l(AE)]²

∴ 900 = 576 + [l(AE)]²

[l(AE)]² = 900 - 576

∴ [l(AE)]² = 324

∴ l(AE) = \(\sqrt{324}\) = 18 m                ... [Taking square root of both sides]

A(Δ ABE) = \(\frac{1}{2}\) × product of sides forming the right angle

= \(\frac{1}{2}\) × l(AE) × l(AB)

= \(\frac{1}{2}\) × 18 × 24

= 216 sq. m

In Δ BCE, a = 30m, b = 28m, c = 26m

Semiperimeter of Δ BCE,

s = \(\frac{a+b+c}{2}\) = \(\frac{30+28+26}{2}\) = \(\frac{84}{2}\)

= 42 m

A (Δ BCE) = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{42(42-30)(42-28)(42-26)}\)

= \(\sqrt{42×12×14×16}\)

= \(\sqrt{7×6×6×2×2×7×8×2}\)

= \(\sqrt{7^2×6^2×2^2×4^2}\)

= 6 × 7 × 2 × 4

= 336 sq. m

In Δ EDC, l(CE) = 28 m, l(DF) = 16 m

A(Δ EDC) = \(\frac{1}{2}\) × base × height

= \(\frac{1}{2}\) × l(CE) × l(DF)

= \(\frac{1}{2}\) ×  28 × 16

= 224 sq. m.

∴ Area of plot ABCDE = A(Δ ABE) + A(Δ BCE) + A(Δ EDC)

= 216 + 336 + 224 = 776 sq. m

∴ The area of the given plot is 776 sq.m.