Important Questions for CBSE Class 9: Area of Parallelograms and Triangles

Very Short Answer Questions [1 Mark Questions]

Ques. State the properties of a parallelogram.

Ans. The three properties of a parallelogram are

• Opposite Sides are equal in length
• The Sum of adjacent angle equals 180°
• The Sum of all interior angles of a parallelogram equals 360°

Ques. State the properties of a triangle.

Ans. The properties of a triangles are

• The Sum of all interior angles is equal to 180°
• The sum of lengths of two sides is greater than the third side.
• Similarly, the difference between two sides of the triangle is less than the third side.

Ques. Two parallelograms are on the same parallels and have the same base. Calculate the area ratio of their respective areas.

Ans. The area of two parallelograms on equal bases and between the same parallels is 1:1.

Ques. In the figure

In the above figure, triangle ABC is proportional to triangle EDC, If AB = 4 cm, ED = 3 cm, CE = 4.2 cm, and CD = 4.8 cm, the CA and CB values are.

1. 6 cm, 6.4 cm
2. 4.8 cm, 6.4 cm
3. 5.4 cm, 6.4 cm
4. 5.6 cm, 6.4 cm

Ans. (d) 5.6 cm, 6.4 cm

Explanation: AB is proportional to ED.

Therefore, AB = x*ED

x = AB/ED = 4/3

Now, CA/CE =4/3, So, CA = 4/3*4.2 = 5.6

Similarly, CB = 6.4

Ques. What will be the area of a parallelogram if its base and height are 4cm and 8cm respectively? 

Ans: Here, b = 4 cm and h = 8 cm 

As we know, 

Area of parallelogram = b x h

= 4 x 8

= 32 sq. cm

Ques. Find the area of the parallelogram if the length of its adjacent sides is 20cm and 25cm and the measure of the angle between the adjacent sides is 34°. 

Ans: Here a = 20cm, b = 25cm, x = 34°

As we know, 

Area of parallelogram = ab sin(x)

= 20 x 25 x sin(34°)

= 500 x 0.55919

= 279.595 cm²

Also read:

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Short Answer Questions [2 Marks Questions]

Ques. ABCD is a parallelogram shown in the below Figure, with AE, DC and CF, AD. Find AD if AB = 16 cm, AE = 8 cm, and CF = 10 cm.

Ans. We are provided with AB = CD = 16 cm where the Opposite sides of these parallelograms are equal. Then, CF = 10 cm and AE = 8 cm

Now let's find the area through the following formula,

Area of parallelogram = Base × height [square. units]

= CD × AE = AD × CF

⇒ 16 × 8 = AD × 10

⇒ AD = 128/10

⇒ AD = 12.8 cm

Ques. XA is a median on side YZ in \(\triangle \text{XYZ}\). Calculate the ratio of ar(XYA) to ar(XZA). Refer the figure below

Ans. We can find that the XA is the median on the side YZ.

∴ YA = AZ

Draw XL is perpendicular to YZ

Then, \(ar (\triangle \text{XYA}) = \frac{1}{2} \times \text{YA} \times \text{XL}\)

\(ar (\triangle \text{XZA}) = \frac{1}{2} \times \text{AZ} \times \text{XL}\)

Thus,, \(ar (\triangle \text{XYA}) : ar (\triangle \text{XZA}) = \frac{1}{2} \times \text{YA} \times \text{XL} : \frac{1}{2} \times \text{AZ} \times \text{XL}\)

The result gives = 1:1 [Therefore, YA = AZ]

Ques. WXYZ is a parallelogram with XP is perpendicular to WZ and ZQ ⊥ WX. If WX = 8 cm, XP = 8 cm and ZQ = 2 cm, find YX. 

Ans. ar(Parallelogram WXYZ) = ar(Parallelogram WXYZ)

WX × ZQ = WZ × XP

8 × 2 = WZ × 8

⇒ WZ = 2 cm

Now, YX = WZ = 2 cm [Hence, the opposite sides of parallelogram are equal]

Ques. In figure, TR is perpendicular to PS, PQ is parallel to TR and PS is also parallel to QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar (PQRS).

Ans. ∴ PQRS is a trapezium where the PQ = 3 cm, QR = 8 cm and SP = 12 cm

Now, TR is parallel to PS and PQ is parallel to TR

∴ PQRT is a rectangle [here the PQ || TR, PT || QR and ∠PTR = 90°]

⇒ PQ = TR = 3 cm

Now, ar(PQRS) = 1/2(PS + QR) × TR = 1/2(12 + 8) × 3 = 30 cm².

Ques. If \(\triangle ABC \sim \triangle PQR\), perimeter of \(\triangle ABC = 32 cm\), perimeter of \(\triangle PQR = 48 cm\) and PR = 6 cm, then find the length of AC. [CBSE, 2012]

Ans. The \(\triangle ABC \sim \triangle PQR\) is given, then

\(\frac{\text{Perimeter of }\triangle ABC}{\text{Perimeter of } \triangle PQR} = \frac{AC}{PR} \implies \frac{32}{48} = \frac {AC}{6} \implies AC=4 cm\)

Also read: Parallelogram Formula

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Long Answer Questions [3 Marks Questions]

Ques. Show that ar (EFGH) = 1/2 ar (EFGH) if E, F, G, and H are the mid-points of the sides of a parallelogram ABCD.

The mid-points of the sides of a parallelogram ABCD are E, F, G, and H, respectively.

To Demonstrate: \(ar (EFGH) = \frac{1}{2} ar (ABCD)\)

H and F are connected together in the construction.

Proof:

\(\frac{1}{2} AD = \frac{1}{2} BC\) (Opposite sides of a ||gm) AD || BC and AD = BC

Furthermore, AH || BF and DH || CF AH = BF and DH = CF (H and F are midpoints)

ABFH and HFCD are hence parallelograms.

EFH and ||gm ABFH, as we know, both lie on the same FH, the common base, and between the same parallel lines AB and HF.

EFH area equals \(\frac{1}{2}\) ABFH area — (i)

Moreover, GHF area equals \(\frac{1}{2}\) HFCD area — (ii)

Adding (i) and (ii), area of EFH + area of GHF = \(\frac{1}{2}\) area of ABFH + \(\frac{1}{2}\) area of HFCD ar (EFGH) = \(\frac{1}{2}\) area of ABFH + \(\frac{1}{2}\) area of HFCD (ABCD).

Ques. ABCD is a parallelogram and Q is any point on side AD. If ar(?QBC) = 10 cm2, find ar(?QAB) + ar(?QDC).

Ans. The parallelograms QBC and ABCD are on the same base BC and sit between the same parallels in this case. The year BC corresponds to the year AD.

∴ ar(QAB) + ar(QDC) + ar(QBC) = 2 ar(QBC) ar(QAB) + ar(QDC) = 2 ar(QBC) ar(QAB) + ar(QDC) = 2 ar(QBC) ar(QAB) + ar(QDC) = 2 ar(QBC) ar(QAB) + ar(QDC) = 2 ar(QBC) ar(QAB) + ar(QDC) = 2 ar(QBC

As a result, ar(QAB) + ar(QDC) = 10 cm2 [ar(QBC) = 10 cm2 (assumed)]

Ques. Calculate the height of an equilateral triangle with side '2a' cm.

Ans. In right angles, the triangle ADB is equal to ADC, then 

AB = AC

AD = AD

\(\therefore \angle ADB = \angle ADC \text{ }(Each = 90 degree)\)

\(\therefore \triangle ADB \cong \triangle ADC (RHS)\)

\(\therefore BD = DC (CPCT)\)

\(\therefore BD = DC = a [\because BC = 2a]\)

In the right \(\triangle ADB, AD^2 + BD^2 = AB^2\) (By Pythagoras Theorem)

\(\implies AD^{2} +a^{2} = (2a)^2\)

\(\implies AD^{2} = 4a^2 - a^{2} = 3a^2\)

\(\implies AD = \sqrt3a \text{ }cm\)

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Very Long Answer Questions [5 Marks Questions]

Ques. ABCD is a quadrilateral (image given below) in which AD = BC and ∠DAB = ∠CBA. Prove that

1. ΔABD ≅ ΔBAC
2. BD = AC
3. ∠ABD = ∠BAC.

Ans. ∠DAB = ∠CBA and AD = BC, as stated in the question.

1. SAS congruence compares ΔABD and ΔBAC, with AB = BA (common arm), ∠DAB = ∠CBA, and AD = BC (given)

As a result, the triangles ABD and BAC are the same, i.e. ΔABD equals ΔBAC. (This is thus demonstrated.)

2. As has already been established, ΔABD ≅ ΔBAC

As a result, BD = AC (by CPCT)

100. Because ΔABD = ΔBAC, the angles are ΔABD = ΔBAC (by Corresponding Parts of Congruent Triangle ).

Ques. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Ans. Given that, AD and BC are two equal perpendiculars to AB.

In order to prove: CD is the bisector of AB

The following Proof is shown below as:

Triangles ΔAOD and ΔBOC are similar and equal by Angle-Angle-Side congruence

Since:

1. ∠A = ∠B (perpendicular angles)
2. AD = BC (given)
3. ∠AOD = ∠BOC (vertically opposite angles)

∴ ΔAOD ≅ ΔBOC.

Therefore, AO = OB ( by Corresponding Parts of Congruent Triangle ).

Thus, the CD bisects AB 

Hence, the problem is proved.

Ques. In the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. [CBSE, 2014]

Ans. From the given question, \(\triangle ADE \text{ }and\text{ } \triangle ABC\),

∠DAE = ∠BAC is similar, and

∠ADE – ∠ABC are the corresponding angles

\(\triangle ADE - \triangle ABC\) Here, this is Angle-Angle Corollary, then

Hence, this is the required solution.