Exercise Set 6.2 Class 9 Maths Ganita Manjari Chapter 6 Perimeter and Area

 

EXERCISE SET 6.2

Measuring Space: Perimeter and Area — Questions & Complete Solutions

1. Question

Find the area of triangle \(ADE\) in Fig. 6.31.

Solution

From the figure, \(AD=BC=8\text{ cm}\). Also, \(DC=10\text{ cm}\), and \(ABCD\) is a rectangle, so the perpendicular distance of \(E\) from \(AD\) is \(10\text{ cm}\).

\[ \begin{aligned} \text{Area}(\triangle ADE) &=\frac{1}{2}\times AD\times 10\\ &=\frac{1}{2}\times8\times10\\ &=40\text{ cm}^{2}. \end{aligned} \]

Answer: \(\boxed{40\text{ cm}^{2}}\)

2. Question

The parallel sides of a trapezium are \(40\text{ cm}\) and \(20\text{ cm}\). If its non-parallel sides are both equal, each being \(26\text{ cm}\), find the area of the trapezium.

Diagram for Q2

Solution

Drop perpendiculars from the ends of the shorter parallel side to the longer parallel side. Since the non-parallel sides are equal, the two right triangles obtained have equal bases.

Difference of parallel sides: \[ 40-20=20\text{ cm}. \] Therefore, each horizontal part is \[ \frac{20}{2}=10\text{ cm}. \]

Let the height be \(h\). By Pythagoras theorem, \[ \begin{aligned} h^2+10^2&=26^2\\ h^2&=676-100=576\\ h&=24\text{ cm}. \end{aligned} \]

\[ \begin{aligned} \text{Area} &=\frac{1}{2}(40+20)\times24\\ &=30\times24\\ &=720\text{ cm}^{2}. \end{aligned} \]

Answer: \(\boxed{720\text{ cm}^{2}}\)

3. Question

Find the area of a triangle, given that its sides are \(8\text{ cm}\) and \(11\text{ cm}\) and \(13\text{ cm}\), and its perimeter is \(32\text{ cm}\).

Solution

Semi-perimeter: \[ s=\frac{32}{2}=16\text{ cm}. \]

By Heron's formula, \[ \begin{aligned} A &=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{16(16-8)(16-11)(16-13)}\\ &=\sqrt{16\times8\times5\times3}\\ &=\sqrt{1920}\\ &=8\sqrt{30}\text{ cm}^{2}. \end{aligned} \]

Answer: \(\boxed{8\sqrt{30}\text{ cm}^{2}}\) (approximately \(43.82\text{ cm}^{2}\))

4. Question

The sides of a triangular plot are in the ratio \(3:5:7\); its perimeter is \(300\text{ m}\). Find its area.

Solution

\[ 3+5+7=15. \] One ratio part is \[ \frac{300}{15}=20\text{ m}. \] Hence the sides are \[ 60\text{ m},\quad100\text{ m},\quad140\text{ m}. \]

Semi-perimeter: \[ s=\frac{60+100+140}{2}=150\text{ m}. \]

By Heron's formula, \[ \begin{aligned} A &=\sqrt{150(150-60)(150-100)(150-140)}\\ &=\sqrt{150\times90\times50\times10}\\ &=1500\sqrt{3}\text{ m}^{2}. \end{aligned} \]

Answer: \(\boxed{1500\sqrt{3}\text{ m}^{2}}\) (approximately \(2598.08\text{ m}^{2}\))

5. Question

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area \(128\text{ cm}^{2}\), find the length of the shorter diagonal.

Diagonals of the rhombus are perpendicular and bisect each other.

Solution

Let the shorter diagonal be \(d\text{ cm}\). Then the longer diagonal is \(2d\text{ cm}\).

Area of a rhombus: \[ A=\frac{1}{2}d_1d_2. \] Therefore, \[ \begin{aligned} 128&=\frac{1}{2}\times d\times2d\\ 128&=d^2\\ d&=\sqrt{128}=8\sqrt2\text{ cm}. \end{aligned} \]

Answer: \(\boxed{8\sqrt2\text{ cm}}\)

6. Question

\(ABCD\) is a parallelogram. \(P\) and \(Q\) are any two points on side \(AB\). What can you say about the ratio area \((\triangle PCD):\) area \((\triangle QCD)\)?

P and Q lie on the same side AB, while both triangles have common base CD.

Solution

Both \(\triangle PCD\) and \(\triangle QCD\) have the same base \(CD\).

Since \(P,Q\) lie on \(AB\) and \(AB\parallel CD\), the perpendicular distances of \(P\) and \(Q\) from \(CD\) are equal to the height \(h\) of the parallelogram.

Thus, \[ \text{Area}(\triangle PCD) =\frac12 CD\cdot h =\text{Area}(\triangle QCD). \]

Answer: \(\boxed{\text{area}(\triangle PCD):\text{area}(\triangle QCD)=1:1}\)

7. Question

\(O\) is any point on the diagonal \(PR\) of a parallelogram \(PQRS\). Prove that the areas of triangles \(PSO\) and \(PQO\) are equal.

O is any point on diagonal PR.

Solution

Since \(O\) lies on diagonal \(PR\), triangles \(PSO\) and \(PSR\) have the same altitude from \(S\) to line \(PR\). Their bases on the same line are \(PO\) and \(PR\).

Therefore, \[ \frac{\text{Area}(\triangle PSO)}{\text{Area}(\triangle PSR)} =\frac{PO}{PR}. \] Similarly, triangles \(PQO\) and \(PQR\) have the same altitude from \(Q\) to line \(PR\), so \[ \frac{\text{Area}(\triangle PQO)}{\text{Area}(\triangle PQR)} =\frac{PO}{PR}. \]

A diagonal of a parallelogram divides it into two equal-area triangles: \[ \text{Area}(\triangle PSR)=\text{Area}(\triangle PQR). \] Hence the two fractions above have equal numerators: \[ \text{Area}(\triangle PSO)=\text{Area}(\triangle PQO). \]

Hence proved: \(\boxed{\text{Area}(\triangle PSO)=\text{Area}(\triangle PQO)}\).

8. Question

If the mid-points of the sides of a 4-gon (also known as a quadrilateral) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

Solution

Let the diagonals of quadrilateral \(ABCD\) intersect at \(O\). Let \(AC=d_1\), \(BD=d_2\), and let the angle between the diagonals be \(\theta\).

Since \(P,Q\) are the midpoints of \(AB,BC\), by the midpoint theorem in \(\triangle ABC\),

\[PQ\parallel AC,\qquad PQ=\frac{1}{2}AC.\]

Similarly, in \(\triangle ADC\),

\[SR\parallel AC,\qquad SR=\frac{1}{2}AC.\]

Also, in \(\triangle BCD\) and \(\triangle BAD\),

\[QR\parallel BD,\qquad QR=\frac{1}{2}BD,\qquad PS\parallel BD,\qquad PS=\frac{1}{2}BD.\]

Thus \(PQ\parallel SR\) and \(QR\parallel PS\). Hence \(PQRS\) is a parallelogram.

Its area is

\[ \begin{aligned} \text{Area}(PQRS) &=PQ\times QR\times\sin\theta\\ &=\frac{d_1}{2}\times\frac{d_2}{2}\times\sin\theta\\ &=\frac{1}{4}d_1d_2\sin\theta. \end{aligned} \]

The area of quadrilateral \(ABCD\), using its diagonals, is

\[\text{Area}(ABCD)=\frac{1}{2}d_1d_2\sin\theta.\]

Therefore,

\[\text{Area}(PQRS)=\frac{1}{2}\text{Area}(ABCD).\]

Hence, \(\boxed{\text{Area of the parallelogram}=\frac{1}{2}\text{ Area of the given quadrilateral}}\).

9. Question

In \(\triangle ABC\), the midpoint of \(BC\) is \(D\) (Fig. 6.32). Median \(AD\) is drawn. \(P\) is any point on \(AD\). Show that area \((\triangle ABP)=\) area \((\triangle ACP)\).

Solution

Since \(D\) is the midpoint of \(BC\), \[ BD=DC. \] Therefore triangles \(ABD\) and \(ACD\) have equal bases \(BD,DC\) and the same altitude from \(A\). Thus, \[ \text{Area}(\triangle ABD)=\text{Area}(\triangle ACD). \]

Now regard \(AD\) as the common base. From the equal areas above, the perpendicular distances of \(B\) and \(C\) from line \(AD\) are equal. Let these equal distances be \(h\).

Since \(P\) lies on \(AD\), both \(\triangle ABP\) and \(\triangle ACP\) have the same base \(AP\), and their heights to \(AP\) are both \(h\). Therefore, \[ \begin{aligned} \text{Area}(\triangle ABP) &=\frac12 AP\cdot h,\\ \text{Area}(\triangle ACP) &=\frac12 AP\cdot h. \end{aligned} \]

Hence, \(\boxed{\text{Area}(\triangle ABP)=\text{Area}(\triangle ACP)}\).

10. Question

Given a square \(ABCD\), let \(P\) be a point within it. Join \(PA,PB,PC,PD\) (Fig. 6.33). What is the ratio of the areas of the red region \((\triangle PAB\text{ and }\triangle PCD)\) and the green region \((\triangle PBC\text{ and }\triangle PDA)\)?

Solution

Let the side of the square be \(a\). Let the perpendicular distances of \(P\) from \(AB,BC,CD,DA\) be \(h_1,h_2,h_3,h_4\), respectively.

Red area: \[ \begin{aligned} A_R &=\text{Area}(\triangle PAB)+\text{Area}(\triangle PCD)\\ &=\frac12ah_1+\frac12ah_3\\ &=\frac12a(h_1+h_3). \end{aligned} \] For opposite parallel sides of a square, \[ h_1+h_3=a. \] Therefore, \[ A_R=\frac12a^2. \]

Green area: \[ \begin{aligned} A_G &=\text{Area}(\triangle PBC)+\text{Area}(\triangle PDA)\\ &=\frac12ah_2+\frac12ah_4\\ &=\frac12a(h_2+h_4). \end{aligned} \] Similarly, \[ h_2+h_4=a, \] so \[ A_G=\frac12a^2. \]

Therefore, \[ \boxed{A_R:A_G=1:1}. \]

11. Question

In \(\triangle ABC\), \(D\) is the midpoint of \(AB\). \(P\) is any point on \(BC\), and \(Q\) is a point on \(AB\) such that \(CQ\parallel PD\). \(PQ\) is joined (Fig. 6.34). Prove that \[ \text{Area}(\triangle BPQ)=\frac12\text{Area}(\triangle ABC). \]

Solution

Since \(D\) is the midpoint of \(AB\), \[ BD=\frac12AB. \] Let \[ \frac{BP}{BC}=t. \]

Because \(PD\parallel CQ\), and \(BP\) and \(BC\) lie on the same straight line while \(BD\) and \(BQ\) lie on \(AB\), triangles \(BPD\) and \(BCQ\) are similar: \[ \triangle BPD\sim\triangle BCQ. \] Therefore, \[ \frac{BQ}{BD}=\frac{BC}{BP}=\frac1t. \] Thus, \[ BQ=\frac{BD}{t}=\frac{AB}{2t}. \]

Let the altitude of \(\triangle ABC\) from \(A\) to \(BC\) be \(h\). Since \(Q\) lies on \(AB\), the perpendicular distance of \(Q\) from \(BC\) is \[ \frac{BQ}{BA}\,h =\frac{1}{2t}h. \]

Now \(\triangle BPQ\) has base \(BP=t\,BC\). Hence, \[ \begin{aligned} \text{Area}(\triangle BPQ) &=\frac12(BP)\left(\frac{h}{2t}\right)\\ &=\frac12(t\,BC)\left(\frac{h}{2t}\right)\\ &=\frac14BC\cdot h. \end{aligned} \] But \[ \text{Area}(\triangle ABC)=\frac12BC\cdot h. \] Therefore, \[ \text{Area}(\triangle BPQ) =\frac12\text{Area}(\triangle ABC). \]

Hence proved: \(\boxed{\text{Area}(\triangle BPQ)=\frac12\text{Area}(\triangle ABC)}\).

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