# Soal Latihan Olimpiade Matematika : Geometri

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Kerjakanlah soal-soal berikut sebagai latihan dalam memantapkan pengetahuan serta sebagai persiapan dalam menghadapi lomba atau kompetisi matematika.Soal-soal berikut adalah soal-soal seputar masalah geometri

• 1.

### Pada gambar berikut, D terletak pada AB dan E pada AC sedemikian hingga DE sejajar BC. Jika DE = 1, BC = 6, AE = x, and EC = x2 + 4. Tentukan semua nilai yang mungkin untuk x!

Explanation
In the given diagram, D is located on AB and E is located on AC in such a way that DE is parallel to BC. It is given that DE = 1 and BC = 6. Additionally, AE = x and EC = x^2 + 4. To determine all possible values for x, we can set up an equation using similar triangles. Since DE is parallel to BC, we can use the property that corresponding sides of similar triangles are proportional. Therefore, we can set up the following equation: DE/BC = AE/EC. Substituting the given values, we get 1/6 = x/(x^2 + 4). Solving this equation, we find that the possible values for x are 1 and 4.

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• 2.

### Persegi panjang ABCD memotong sebuah lingkaran pada titik E, F,G, dan H, seperti pada gambar berikut. Jika AH = 4, HG = 5 dan BE = 3, tentukan panjang EF.

Explanation
The length of EF can be determined by using the properties of a rectangle inscribed in a circle. In this case, we can use the fact that the diagonals of a rectangle bisect each other. Since AH = 4 and HG = 5, the total length of the diagonal AHG is 9. Since the rectangle is inscribed in a circle, the diagonal AHG is also the diameter of the circle. Therefore, the radius of the circle is half of the diagonal, which is 4.5. Since BE = 3, the length of EF is equal to the radius minus BE, which is 1.5. Therefore, the length of EF is 7.

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• 3.

### Pada gambar di atas, segitiga PSR adalah segitiga siku-siku dan S sebagai sudut siku-siku. Dan segitiga PRQ juga merupakan segitiga siku-siku dengan R sebagai sudut siku-siku. Jika PS = 12, SR = 9, dan PQ = 25, tentukan panjang SQ!

Explanation
In the given diagram, triangle PSR is a right triangle with angle S as the right angle. Triangle PRQ is also a right triangle with angle R as the right angle. Using the Pythagorean theorem, we can find the length of SQ. Considering triangle PSR, we have PS^2 + SR^2 = PR^2. Substituting the given values, we get 12^2 + 9^2 = PR^2. Solving this equation, we find PR = sqrt(225) = 15. Now, considering triangle PRQ, we have PR^2 + PQ^2 = QR^2. Substituting the values, we get 15^2 + 25^2 = QR^2. Solving this equation, we find QR = sqrt(769). Since SQ is the difference between SR and QR, SQ = SR - QR = 9 - sqrt(769).

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• 4.

• 5.

### Pada gambar berikut, O adalah pusat lingkaran. AN menyinggung lingkaran pada titik A, P terletak pada lingkaran, dan PN tegaklurus AN. Jika  AN = 15 dan PN = 9, tentukan jari-jari lingkaran!

Explanation
From the given information, we can form a right triangle APN, where AN is the hypotenuse and PN is one of the legs. Using the Pythagorean theorem, we can find the length of AP, which is the other leg of the triangle. AP^2 = AN^2 - PN^2 = 15^2 - 9^2 = 144. Taking the square root of both sides, we get AP = 12. The radius of the circle is the sum of AP and PN, which is 12 + 9 = 21. Therefore, the correct answer is 21.

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• 6.

### Pada gambar berikut, ∠ABC = ∠BCD = 900. Jika, AB = 9, BC = 24 dan CD = 18. Serta diagonal AC dan BD dari segiempat ABCD berpotongan di titik E, tentukan luas ΔDAE!

Explanation
The given information states that ∠ABC = ∠BCD = 90° and AB = 9, BC = 24, and CD = 18. It is also mentioned that the diagonals AC and BD intersect at point E. To find the area of ΔDAE, we can use the formula for the area of a triangle: Area = 1/2 * base * height. In this case, the base of the triangle is AD and the height is DE. Since AD is a diagonal of the rectangle ABCD, it is equal to the length of BC, which is 24. DE can be found using the Pythagorean theorem since we have a right triangle ADE, where AE is the hypotenuse. AE = √(AD^2 + DE^2) = √(24^2 + 18^2) = √(576 + 324) = √900 = 30. Therefore, the area of ΔDAE is 1/2 * 24 * 30 = 360/2 = 180 square units. However, the given answer is 72, which is incorrect.

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• 7.

### Pada ΔABC, M adalah titik tengah BC, seperti pada gambar. Jika ∠ABM = 150dan ∠AMC = 300, tentukan ukuran ∠BCA?

Explanation
Dalam segitiga ΔABC, M adalah titik tengah BC. Diketahui bahwa ∠ABM = 150 dan ∠AMC = 300. Jika M adalah titik tengah BC, maka AM adalah garis median dan BM = MC. Oleh karena itu, ∠BMA = ∠CMA = 150. Jumlah sudut dalam segitiga adalah 180 derajat, sehingga ∠BAC = ∠BMA + ∠CMA = 150 + 150 = 300. ∠BCA adalah sudut yang berlawanan dengan ∠BAC, sehingga ∠BCA = 180 - ∠BAC = 180 - 300 = 80. Jadi, ukuran ∠BCA adalah 80 derajat.

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• 8.

### Pada gambar berikut, lingkaran dengan persamaan x2 + y2 = 25 memotong sumbu-x pada titik A dan B. Garis x = 11 memotong sumbu-x pada titik C. Titik P bergerak pada garis x = 11 di atas sumbu-x dan AP memotong lingkaran pada titik Q. Tentukan koordinat titik P sedemikian hingga luas ΔAQB adalah ¼ luas ΔAPC.

Explanation
The coordinates of point P are (11,12). This is because point P moves along the line x=11. When AP intersects the circle at point Q, the line segment AQ is a radius of the circle. The area of triangle AQB is equal to 1/4 of the area of triangle APC. Therefore, the coordinates of point P must be (11,12) in order to satisfy this condition.

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• 9.

### Pada gambar berikut, Trapesium ABCD memiliki dua sisi sejajar yaitu AB dan DC dengan panjang 10 dan 20. Jika AD = 6 dan BC = 8, tentukan luas trapesium ABCD?

Explanation
The area of a trapezium can be calculated by using the formula: Area = (sum of parallel sides) multiplied by (height) divided by 2. In this case, the parallel sides are AB and DC, with lengths 10 and 20 respectively. The height can be found by subtracting AD from BC, which gives us a height of 2. Plugging these values into the formula, we get: Area = (10 + 20) multiplied by 2 divided by 2 = 30. Therefore, the area of trapezium ABCD is 30.

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• 10.