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  • 2 answers

Avisha Joseph 5 years, 5 months ago

1

Richa Shree 5 years, 5 months ago

-1
  • 2 answers

Gaurav Seth 5 years, 5 months ago


Given a circle with centre O and chords AB = CD
Draw OP⊥ AB and OQ ⊥ CD
Hence AP = BP = (1/2)AB and CQ = QD = (1/2)CD
Also ∠OPA = 90° and ∠OQC = 90°
Since AB = CD
⇒ (1/2) AB = (1/2) CD
⇒ AP = CQ
In Δ’s OPA and OQC,
∠OPA = ∠OQC = 90°
AP = CQ (proved)
OA = OC (Radii)
∴ ΔOPA ≅ ΔOQC (By RHS congruence criterion)
Hence OP = OQ (CPCT)

Rohit Kumar 5 years, 5 months ago

Kise birat ki bara ber jeva kendra par baraber kod antertit karti ha
  • 5 answers

Varun Jakhar 5 years, 5 months ago

22/7 or 3.14

Ashutosh Gupta 5 years, 5 months ago

22/7

Ramchand Babu Ps 5 years, 5 months ago

3.14

Pushp Raj 5 years, 5 months ago

22 upon 7

Rudrani Kumari 5 years, 5 months ago

22/7
  • 1 answers

Rohit Singh 5 years, 5 months ago

In parallelogram sum of adjacent angle are 180 so x+y=180
  • 0 answers
  • 1 answers

Kajal Yadav 5 years, 5 months ago

With hands
  • 3 answers

Diya Tomar 5 years, 5 months ago

Median formula for odd number is n+1÷2 (is total number of odd observation 7+1÷2=4 so median is 4th number that is 3

Avisha Joseph 5 years, 5 months ago

6

Ronit Bhatia 5 years, 5 months ago

3
  • 0 answers
  • 1 answers

Partap Gupta 5 years, 5 months ago

The line segment joining the mid points of two sides of a triangle is parallel to third side.
  • 0 answers
  • 0 answers
  • 6 answers

Riddhima Rewaria 5 years, 5 months ago

πrl

Rohit Singh 5 years, 5 months ago

πrl

Anish Bhuria 5 years, 5 months ago

πrl

Ruchi Pal 5 years, 5 months ago

CSA of cone is πrl

Ankit Roy 5 years, 5 months ago

Πrl

E Verma 5 years, 5 months ago

πrl
  • 1 answers

Nirmal Choudhary 5 years, 5 months ago

,xzxx
  • 0 answers
  • 1 answers

Yogita Ingle 5 years, 5 months ago

a-b=7 ⇒a=7+b .....(i)
a²+b²=85  ....(ii)

Substitution of equation 1 to equation 2.
(7+b)²+b² =85
49+14b +b²+b² =85
2b²+14b-36 =0
  b² +7b -18 =0
 (b+9)(b-2) =0
b=-9 v b=2

Find value of a
a-b=7
⇒if b=-9
a-(-9)=7
a =7-9
   =-2
⇒if b=2
a-2=7
a = 7+2
   =9

So value of a³ - b³ there are 2 options
⇒if (a,b) = (-2,-9)
a³-b³ = -2³ - (-9)³
         = -8 +729
         = 721

⇒if (a,b) = (2,9)
a³-b³ = 2³ - 9³  
         = 8-729
         = -721

  • 1 answers

Yogita Ingle 5 years, 5 months ago

a-b=7 ⇒a=7+b .....(i)
a²+b²=85  ....(ii)

Substitution of equation 1 to equation 2.
(7+b)²+b² =85
49+14b +b²+b² =85
2b²+14b-36 =0
  b² +7b -18 =0
 (b+9)(b-2) =0
b=-9 v b=2

Find value of a
a-b=7
⇒if b=-9
a-(-9)=7
a =7-9
   =-2
⇒if b=2
a-2=7
a = 7+2
   =9

So value of a³ - b³ there are 2 options
⇒if (a,b) = (-2,-9)
a³-b³ = -2³ - (-9)³
         = -8 +729
         = 721

⇒if (a,b) = (2,9)
a³-b³ = 2³ - 9³  
         = 8-729
         = -721
 

  • 0 answers
  • 2 answers

Smita Malvankar 5 years, 5 months ago

Hi

Yogita Ingle 5 years, 5 months ago

Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m.

Join the diagonal BD which divides quadrilateral ABCD in two triangles i.e ∆BCD & ∆ABD.

In ΔBCD,
By applying Pythagoras Theorem

BD²=BC² +CD²

BD²= 12²+ 5²= 144+25

BD²= 169
BD = √169= 13m

∆BCD is a right angled triangle.

Area of ΔBCD = 1/2 ×base× height

=1/2× 5 × 12= 30 m²

For ∆ABD,
Let a= 9m, b= 8m, c=13m

Now,
Semi perimeter of ΔABD,(s) = (a+b+c) /2

s=(8 + 9 + 13)/2 m
= 30/2 m = 15 m

s = 15m

Using heron’s formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 – 9) (15 – 9) (15 – 13)

= √15 × 6 × 7× 2
=√5×3×3×2×7×2
=3×2√35
= 6√35= 6× 5.92

[ √6= 5.92..]
= 35.52m² (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD
= 30+ 35.5= 65.5 m²

Hence, area of the park is 65.5m²

  • 0 answers
  • 0 answers
  • 1 answers

Dipika Pal 5 years, 5 months ago

If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment the four points lie on a circle

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