ML Aggarwal Solution Class 9 Chapter 11 Mid Point Theorem MCQS

  MCQS

Choose the correct answer from the given four options (1 to 6):

Question 1

In a ∆ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm. IfD and E are mid-points of AB and BC respectively, then the length of DE is

(a) 1.5 cm

(b) 2 cm

(c) 2.5 cm

(d) 3.5 cm

Sol :

In ∆ABC, D and E are the mid-points of sides AB and BC











∴DE$=\frac{1}{2}$AC

But AC=5 cm

∴DE$=\frac{1}{2}\times 5=\frac{5}{2}$ cm =2.5cm

Ans (c)


Question 2

In the given figure, ABCD is a rectangle in which AB = 6 cm and AD = 8 cm. If P and Q are mid-points of the sides BC and CD respectively, then the length of PQ is

(a) 7 cm

(b) 5 cm

(c) 4 cm

(d) 3 cm

Sol :
In the given figure
ABCD is a rectangle AB=6 cm , AD=8 cm
P and Q are mid points of BC and CD
∴PQ$=\frac{1}{2}$BD

But BD$=\sqrt{BC^2+CD^2}$ (pythagoras theorem)






$=\sqrt{8^2+6^2}$

$=\sqrt{64+36}=\sqrt{100}$

=10 cm

∴PQ$=\frac{1}{2}$BD

$=\frac{1}{2}\times 10=5$ cm

Ans (b)


Question 3

D and E are mid-points of the sides AB and AC of ∆ABC and O is any point on the side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, then DEQP is

(a) a square

(b) a rectangle

(c) a rhombus

(d) a parallelogram

Sol :

D and E are mid-points of sides AB and AC respectively of AABC O is any point on BC and AO is joined P and Q are mid-points of OB and OC, EQ and DP are joined











∵DE$=\frac{1}{2}BC $ and DE||BC..(i)

∴P and Q are mid points of BO and OC

∴PQ=PO+OQ

$=\frac{1}{2}$BO+$\frac{1}{2}$OC

$=\frac{1}{2}$(BO+OC)

$=\frac{1}{2}$(BC)..(ii)

=DE

From EQPD is a parallelogram 


Question 4

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle if

(a) PQRS is a parallelogram

(b) PQRS is a rectangle

(c) the diagonals of PQRS are perpendicular to each other

(d) the diagonals of PQRS are equal.

Sol :

A, B, C and D are the mid-points of the sides PQ, QR, RS and SP respectively











The quadrilateral so formed by joining the mid points A, B , C , D is ABCD

ABCD will be rectangle , if the diagonals of PQRS bisects each other

i.e. PR and QS bisects each other 

Ans (c)


Question 5

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rhombus if

(a) ABCD is a parallelogram

(b) ABCD is a rhombus

(c) the diagonals of ABCD are equal

(d) the diagonals of ABCD are perpendicular to each other.

Sol :

P, Q, R and S are the mid-points of the quadrilateral ABCD and a quadrilateral is formed by joining the mid-points in order











PQRS will be rhombus if the diagonals of ABCD are equal

i.e. AC=BD 

Ans (c)


Question 6

The figure formed by joining the mid points of the sides of a quadrilateral

ABCD, taken in order, is a square only if

(a) ABCD is a rhombus r

(b) diagonals of ABCD are equal

(c) diagonals of ABCD are perpendicular to each other

(d) diagonals of ABCD are equal and perpendicular to each other.

Sol :
P,Q,R and S are the mid points of the quadrilateral ABCD and a quadrilateral is formed by joining them in order. The quadrilateral so formed will be a square if the diagonals of ABCD are equal and perpendicular to each other.












i.e. AC and BD are equal and bisect it perpendicular 

Ans (d)

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more