Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9
AI Mentor
Check Your IQ
Free Expert Demo
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course
Offline Centres

Q.

A figure of an aeroplane is shown below:


Question Image

Here, the wings ABCD and FGHI   are parallelograms, the tail DEF   is an isosceles triangle, the cockpit CKI   is a semicircle and CDFI   is a square.


Also, BP⊥CD  , HO⊥FI   and EL⊥DF  .


If CD=8 cm,BP=HQ=4cm and DE=EF=5cm  , then the area of the whole figure is ____ cm 2  . [Takeπ=3.14.]  


see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

A figure of an aeroplane is shown below:
Question ImageHere, the wings ABCD and FGHI   are parallelograms, the tail DEF   is an isosceles triangle, the cockpit CKI   is a semicircle and CDFI   is a square.
Also, BP⊥CD  , HO⊥FI   and EL⊥DF  .
If CD=8 cm,BP=HQ=4cm and DE=EF=5cm  , then the area of the whole figure is 165.12 cm 2  .
The area of parallelogram is given as-
Area=Base×Height  
Now, Area of the parallelogram ABCD=8×4=32  cm 2  
Area of the parallelogram FGHI=8×4=32  cm 2  
Area of the square CDFI=8×8=64  cm 2  
The pythagoras theorem for a right angled-triangle is given as- Hypotenuse 2 = Perpendicular 2 + Base 2   Using Pythagoras theorem for ΔELF  , we get
⇒E L 2 = 5 2 − 4 2 ⇒EL= 25−16 ⇒EL= 9 ⇒EL=3 cm  
The area of a triangle is given as-
  Area= 1 2 ×Base×Height  
The area of a semicircle with radius r   is given as-
Area= 1 2 π r 2  
  Area of ΔDEF= 1 2 ×8×3=12  cm 2  
Area of the semicircle CKI= 1 2 × 22 7 ×16×16=25.12  cm 2  
Therefore, the total area is-
Total area =32+32+64+12+25.12=165.12  cm 2  
Hence, 165.12  cm 2   is the required area.
 
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring