Jharkhand Board JAC Class 9 Maths Important Questions Chapter 9 Areas of Parallelograms and Triangles Important Questions and Answers.

## JAC Board Class 9th Maths Important Questions Chapter 9 Areas of Parallelograms and Triangles

Question 1.

In a parallelogram ABCD; AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 cm and 5 cm. Find AD.

Solution :

We know that, Area of a parallelogram

= Base × Corresponding altitude

∴ Area of parallelogram

ABCD = AD × BN = AB × DM

⇒ AD × 5 = 8 × 4

⇒ AD = \(\frac{8 \times 4}{5}\) = 6.4 cm.

Question 2.

ABCD is a quadrilateral and BD is one of its diagonals as shown in the figure. Show that the quadrilateral ABCD is a parallelogram and find its area.

Solution :

From figure, the transversal DB is intersecting a pair of lines DC and AB such that

∠CDB = ∠ABD = 90°.

As these angles form a pair of alternate interior angles

∴ DC || AB.

Also, DC = AB = 2.5 units.

∴ Quadrilateral ABCD is a parallelogram. Now, area of parallelogram ABCD

= Base × Corresponding altitude

= 2.5 × 4 = 10 sq. units

Question 3.

In figure, E is any point on median AD of a ΔABC. Show that ar(ABE) = ar(ACE).

Solution :

Construction: From A, draw AG ⊥ BC and from E, draw EF ⊥ BC.

Proof: ar(ΔABD) = \(\frac {BD × AG}{2}\)

ar(ΔADC) = \(\frac{\mathrm{DC} \times \mathrm{AG}}{2}\)

But, BD = DC [∵ D is the mid-point of BC as AD is the median]

∴ ar(ΔABD) = ar(ΔADC) ……………(i)

Again, ar(ΔEBD) = \(\frac{\mathrm{BD} \times \mathrm{EF}}{2}\)

ar(ΔEDC) = \(\frac{\mathrm{DC} \times \mathrm{EF}}{2}\)

But, BD = DC

∴ ar(ΔEBD) = ar(ΔEDC) …….(ii)

Subtracting (ii) from (i), we get

ar(ΔABD) – ar(ΔEBD)

= ar(ΔADC) – ar(ΔEDC)

⇒ ar(ΔABE) = ar(ΔACE)

Hence, proved.

Question 4.

Triangles ABC and DBC are on the same base BC; with A, D on opposite sides of the line BC, such that ar(ΔABC) = ar(ΔDBC). Show that BC bisects AD.

Solution :

Construction: Draw AL ⊥ BC and DM ⊥ BC.

Proof: ar(ΔABC) = ar(ΔDBC) (Given)

⇒ \(\frac{\mathrm{BC} \times \mathrm{AL}}{2}=\frac{\mathrm{BC} \times \mathrm{DM}}{2}\)

⇒ AL = DM ………….(i)

Now in Δs OAL and OMD

AL = DM [From (i)]

⇒ ∠ALO = ∠DMO [Each = 90°]

⇒ ∠AOL = ∠MOD [Vert. opp. ∠s]

∴ ΔOAL ≅ ΔODM [By AAS]

∴ OA = OD [By CPCT]

i.e., BC bisects AD. Hence, proved.

Question 5.

ABC is a triangle in which D is the mid-point of BC and E is the mid-point of AD. Prove that the area of ΔBED = \(\frac {1}{4}\) area of ΔABC.

Solution :

Given: A ΔABC in which D is the midpoint of BC and E is the mid-point of AD.

To prove: ar(ΔBED) = \(\frac {1}{4}\)ar(ΔABC).

Proof: ∵ AD is a median of ΔABC

∴ ar(ΔABD) = ar(ΔADC)

= \(\frac {1}{2}\)ar(ΔABC) ……..(i)

[∵ Median of a triangle divides it into two triangles of equal area]

Again,

∵ BE is a median of ΔABD.

∴ ar(ΔBEA) = ar(ΔBED)

= \(\frac {1}{2}\)ar(ΔABD) ……..(ii)

[∵ Median of a triangle divides it into two triangles of equal area]

∴ ar(ΔBED) = \(\frac {1}{2}\)ar(ΔABD)

= \(\frac {1}{2}\) × \(\frac {1}{2}\)ar(ΔABC) [From (i)]

∴ ar(ΔBED) = \(\frac {1}{4}\)a(ΔABC). [From (ii)]

Question 6.

Prove that the area of an equilateral triangle is equal to \(\frac{\sqrt{3}}{4}\)a^{2}, where a is the side of the triangle.

Solution :

Draw AD ⊥ BC

⇒ ΔABD ≅ ΔACD

[By RHS congruence rule]

∴ BD = DC [By CPCT]

∴ BD = DC = \(\frac {a}{2}\)

In right-angled ΔABD

AD^{2} = AB^{2} – BD^{2}

= a^{2} – (\(\frac {a}{2}\))^{2} = a^{2} – \(\frac{a^2}{4}=\frac{3 a^2}{4}\)

AD = \(\frac{\sqrt{3} a}{2}\)

Area of ΔABC = \(\frac {1}{2}\)BC × AD

= \(\frac {1}{2}\)a × \(\frac{\sqrt{3} a}{2}=\frac{\sqrt{3} a^2}{4}\)

Hence, proved.

Question 7.

In figure, P is a point in the interior of rectangle ABCD. Show that

(i) ar(ΔAPD) + ar(ΔBPC)

= \(\frac {1}{2}\)ar(rect. ABCD)

(ii) ar(APD) + ar(PBC)

= ar(APB) + ar(PCD)

Solution :

Given: A rect. ABCD and P is a point inside it. PA, PB, PC and PD have been joined.

To prove:

(i) ar(ΔAPD) + ar(ΔBPC)

= \(\frac {1}{2}\)ar(rect. ABCD)

(ii) ar(ΔAPD) + ar(ΔBPC)

= ar(ΔAPB) + ar(ΔCPD).

Construction :

Draw EPF || AB

and LPM || AD.

Proof: EPF || AB and DA cuts them,

(i) ∠EAB = 90° [As each angle of a rectangle is of 90°]

∴ ∠DEP = ∠EAB = 90° [Corresponding angles]

∴ PE ⊥ AD

Similarly, PF ⊥ BC; PL ⊥ AB and PM ⊥ DC.

∴ ar(ΔAPD) + ar(ΔBPC)

= (\(\frac {1}{2}\) × AD × PE) + ar (\(\frac {1}{2}\) × BC × PF)

= \(\frac {1}{2}\)AD (PE + PF) [∵ BC = AD]

= \(\frac {1}{2}\) × AD × EF = \(\frac {1}{2}\) × AD × AB [∵ EF = AB]

= \(\frac {1}{2}\) × ar(rectangle ABCD)

(ii) ar(ΔAPB) + ar(PCD)

= (\(\frac {1}{2}\) × AB × PL) + (\(\frac {1}{2}\) × DC × PM)

= \(\frac {1}{2}\) × AB × (PL + PM) [∵ DC = AB]

= \(\frac {1}{2}\) × AB × LM

= \(\frac {1}{2}\) × AB × AD [∵ LM = AD]

= \(\frac {1}{2}\) × ar(rect. ABCD).

= ar(ΔAPD) + ar(PBC)

= ar(ΔAPB) + ar(PCD)

Hence, proved.

Multiple Choice Questions

Question 1.

The sides BA and DC of the parallelogram ABCD are produced as shown in the figure then

(a) a + x = b + y

(b) a + y = b + a

(c) a + b = x + y

(d) a – b = x – y

Solution :

(c) a + b = x + y

Question 2.

The sum of the interior angles of polygon is three times the sum of its exterior angles. Then numbers of sides in polygon is

(a) 6

(b) 7

(c) 8

(d) 9

Solution :

(d) 9

Question 3.

In the following figure, AP and BP are angle bisectors of ∠A and ∠B which meet at a point P of the parallelogram ABCD. Then 2∠APB =

(a) ∠A + ∠B

(b) ∠A + ∠C

(c) ∠B + ∠D

(d) 2∠C + ∠D

Solution :

(a) ∠A + ∠B

Question 4.

In a parallelogram ABCD, AO and BO are respectively the angle bisectors of ∠A and ∠B (see figure). Then measure of ∠AOB is

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Solution :

(d) 90°

Question 5.

In a parallelogram ABCD, ∠D = 60° then the measurement of ∠A is

(a) 120°

(b) 65°

(c) 90°

(d) 75°

Solution :

(a) 120°

Question 6.

In the adjoining figure ABCD, the angles x and y are

(a) 60°, 30°

(b) 30°, 60°

(c) 45°, 45°

(d) 90°, 90°

Solution :

(a) 60°, 30°

Question 7.

In the parallelogram PQRS (see figure), the values of ∠SQP and ∠QSP are

(a) 45°, 60°

(b) 60°, 45°

(c) 70°, 35°

(d) 35°, 70°

Solution :

(a) 45°, 60°

Question 8.

In parallelogram ABCD, AB = 12 cm. The altitudes corresponding to the sides CD and AD are respectively 9 cm and 11 cm. Find AD.

(a) \(\frac {108}{11}\) cm

(b) \(\frac {108}{10}\) cm

(c) \(\frac {99}{10}\) cm

(d) \(\frac {108}{17}\) cm

Solution :

(a) \(\frac {108}{11}\) cm

Question 9.

In ΔABC, AD is a median and P is a point on AD such that AP : PD = 1 : 2 then the area of ΔABP =

(a) \(\frac {1}{2}\) × Area of ΔABC

(b) \(\frac {2}{3}\) × Area of ΔABC

(c) \(\frac {1}{3}\) × Area of ΔABC

(d) \(\frac {1}{6}\) × Area of ΔABC

Solution :

(d) \(\frac {1}{6}\) × Area of ΔABC

Question 10.

In ΔABC if D is a point on BC and divides it in the ratio 3 : 5 i.e., if BD : DC = 3 : 5 then, ar(AADC): ar(ΔABC) = ?

(a) 3 : 5

(b) 3 : 8

(c) 5 : 8

(d) 8 : 3

Solution :

(b) 3 : 8