prepare question and solution of quadrilateral
Answers
Step-by-step explanation:
Question 1:
Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠∠XYZ = 135∘135° then what is the measure of ∠∠XWZ and ∠∠YZW ?
If l(OY)= 5 cm then l(WY)= ?
ANSWER:
In a parallelogram, opposite angles are congruent.
∠∠XYZ = ∠∠XWZ = 135∘135°
Also, WX || ZY
So, ∠∠YZW + ∠∠XWZ = 180°° (Since, interior angles on the same side of the transversal are supplementary)
⇒∠YZW+135°=180°⇒∠YZW=45°⇒∠YZW+135°=180°⇒∠YZW=45°
Now l(OY)= 5
we know that diagonals of a parallelogram bisect each other.
So, l(WY) = 2×OY=2×5=10 cm2×OY=2×5=10 cm
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Question 2:
In a parallelogram ABCD, If ∠∠A =(3x + 2)∘3x + 2° , ∠∠B = (2x − 32)∘2x - 32° then find the value of x and then find the measures of ∠∠C and ∠∠D.
ANSWER:
Interior angles on the same side of the transversal are supplementary.
∠∠A +∠∠B = 180°°
⇒⇒(3x + 12)∘3x + 12° + (2x − 32)∘2x - 32° = 180°°
⇒5x−20=180°⇒5x=200⇒x=40°⇒5x-20=180°⇒5x=200⇒x=40°
Thus, ∠∠A = 3×40°+12=132°3×40°+12=132°
∠B=2x−32=2×40°−32=48°∠B=2x-32=2×40°-32=48°
Also, opposite angles of a parallelogram are congruent.
∠A=∠C=132°∠B=∠D=48°∠A=∠C=132°∠B=∠D=48°
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Question 3:
Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.
ANSWER:
Perimeter = 150 cm
Let one of the sides be x.
Other side = x+25x+25
Perimeter = Sum of all sides
x+x+(x+25)+(x+25)=150⇒4x+50=150⇒4x=100⇒x=25x+x+x+25+x+25=150⇒4x+50=150⇒4x=100⇒x=25
Thus, the sides are 25 cm, 25 cm, 50 cm and 50 cm
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Question 4:
If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.
ANSWER:
Let the two adjacent angles be x and 2xx and 2x.
Interior angles on the same side of the transversal are supplementary.
So,
x+2x=180°⇒3x=180°⇒x=60°x+2x=180°⇒3x=180°⇒x=60°
Thus, the adjacent angles are 60° and 120°60° and 120°.
The opposite angles of a parallelogram are congruent.
So, the angles of the parallelogram will be 60°,120°60°,120°, 60° and 120°60° and 120°.
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Question 5:
Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that □□ABCD is a rhombus.
ANSWER:
In △△AOB,
AO = 5 cm
OB = 12 cm
AB = 13 cm
5, 12 and 13 form a Pythagorean triplet.
Thus, △△AOB is a right angle triangle, right angled at O.
ABCD is a parallelogram.
So, diagonals bisect each other.
⇒⇒AO = OC = 5 cm
Also, OB = OD = 12 cm.
Thus, in parallelogram ABCD, diagonals bisect at right angles.
Hence, ABCD is a rhombus.
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Question 6:
In the given figure, □□PQRS and □□ABCR are two parallelograms. If ∠∠P = 110∘110° then find the measures of all angles of □□ABCR.
ANSWER:
In □□PQRS,
∠∠P = 110∘110°
We know that opposite angles of a parallelogram are congruent.
So, ∠∠P = ∠∠R = 110∘110°
In □□ABCR,
∠∠R =∠∠B = 110∘110° (Opposite angles of a parallelogram are congruent)
Interior angles on the same side of the transversal are supplementary.
So,
∠R+∠A=180°⇒110°+∠A=180°⇒∠A=180°−110°⇒∠A=70°∠R+∠A=180°⇒110°+∠A=180°⇒∠A=180°-110°⇒∠A=70°
And ∠∠A =∠∠C = 110°°.
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Question 7:
In the given figure, □□ABCD is a parallelogram. Point E is on the ray AB such that BE = AB then prove that line ED bisects seg BC at point F.
ANSWER:
□□ABCD is a parallelogram
So, AD || CB
⇒⇒AD || BF
Given: AB = BE so, B is the midpoint of AE.
By converse of mid-point theorem,
F is the midpoint of DE. So, DF = FE
In △△EBF and △△DCF,
DF = FE (Proved above)
DC = BE (Since DC = AB and AB = BE)
∠∠FDC = ∠∠FEB (Alternate interior angles of the parallel lines AE and CD)
Thus, △△EBF ≅≅ △△DCF (SAS congruency)
Therefore, FB = FC (CPCT)
Hence, ED bisects seg BC at F.
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Question 1:
In the given figure, □□ABCD is a parallelogram, P and Q are midpoints of side AB and DC respectively, then prove □□APCQ is a parallelogram.
ANSWER:
□□ABCD is a parallelogram,
AB ≅≅ CD and AB || CD
P and Q are the mid points of AB and CD respectively.
So, AP ≅≅ CQ and AP || CQ
Thus, □□APCQ is also a parallelogram.
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Question 2:
Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.
ANSWER:
Let ABCD be a rectangle.
∠∠A = ∠∠B = ∠∠C = ∠∠D = 90°90°.
For any quadrilateral to be a parallelogram, pair of opposite angles should be congruent.
In rectangle ABCD,
∠∠A = ∠∠C = 90°90°
∠∠B = ∠∠D = 90°90°
Thus, rectangle ABCD is a paralleogram.
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Question 3:
In the given figure, G is the point of concurrence of medians of Δ∆DEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that
□□GEHF is a parallelogram.
ANSWER:
G is the point of concurrence of the medians of Δ∆DEF.
Let the point where the median divides EF into two equal parts be A.
Thus, EA = AF. .....(1)
we know that the point of concurrence of the medians, divides each median in the ratio 2 : 1.
So, let DG = 2x and GA = x
Given that DG = GH
So, GA = AH = x
Thus, point A dividess EF and GH into two equal parts.
Hence, □□GEHF is a parallelogram as the diagonals EF and GH bisect each other.