Question

Q5.Pinki and Rinki are sisters. They go from their house A along a straight path and reaches a toy shop D. D is along the straight road BC. After buying gifts Pinki went to her friend's house B and Rinki to her friend's house C for birthday party. B and C are at equal distance from toy shop. Area of triangle ADC is 36 Sq m and altitude from A to BC is half of DC. Can you tell the: A a) Area of triangle ABD b) Length of altitude from A to BC c) Area of triangle ABC


Please answer and it is not a MCQ ​

Answer 1

Hope it helps.

Please mark it as brainliest!

Answer 2

Given:

Area of triangle ΔADC = 36 m²

$Altitude~AD=\cfrac{1}{2} ~DC$

D(Toy shop) is at an equal distance from B and C.

To Find:

Area of triangle ΔABD.

Area of triangle ΔABC.

Length of Altitude AD.

Solution:

                                            A

                                           /|\

                                         /  |   \

                                       /    |     \

                                B  /       |  $^{\underline{36}}$  \  C

                                            D

For Finding the Area of ΔABD,

The ΔABD and ΔADC should be proved congurent. By using the congruency rules,

                                BD = DC   {∴ Given

                                AD = AD  {∴ Common Line

                           ∠ABD = ∠ADC  {∴ Both are of 90°

So, the ΔABD and ΔADC are congruent by the SAS congruency rule.

                             ΔABD ≅ ΔADC

Now, It is known that the congruent triangles are equal in area.

Area of ΔABD  = Area of ΔADC

                        = 36 m²

Hence the Area of ΔABD is 36 m².

Since the Area of ΔABC is equal to the sum of the Area of Δs ABD and ADC.

Area of ΔABC = Area of ΔABD + Area of ΔADC

                        = 36 +36

                        = 72 m²

Hence the Area of ΔABC is 72 m².

It is given that Altitude AD is equal to half of DC.

                         $Altitude~AD=\cfrac{1}{2} ~DC$

Now, From the ΔADC,

                    $Area~of~\triangle ADC = \cfrac{1}{2}*b*h\\~~~~~~~~~~~~~~~~~~~~36=\cfrac{1}{2}*DC*\cfrac{1}{2}~DC\\~~~~~~~~~~~~~~~~~~~~36=\cfrac{1}{4}*DC^2\\~~~~~~~~~~~~~~~~~DC^2= 144\\~~~~~~~~~~~~~~~~~~DC=1 2~m$

Using the value of DC = 12 m,

                           $Altitude~AD=\cfrac{1}{2} *12\\Altitude~AD=\c6$

Hence the value of Altitude AD is 6 m.